CODEFORCES-PROBLEMSET
水题 然而看不仔细爆int了
c++
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
double n, m, a;
cin >> n >> m >> a;
ll ans = ll(ceil(n / a)) * ll(ceil(m / a));
cout << ans << endl;
return ;
}
py3
import math
t = input().split()
n, m, a = map(int, t)
print(int(math.ceil(n / a) * math.ceil(m / a)))
有毒的模拟水题......
#include <bits/stdc++.h>
using namespace std; int main(){
int n;
cin >> n;
while(n--){
string s;
cin >> s;
bool flag = false;
int i = ;
while(i < s.size() && 'A' <= s[i] && s[i] <= 'Z') ++i;
for(; i < s.size(); ++i){
if('A' <= s[i] && s[i] <= 'Z'){
flag = true;
break;
}
}
if(flag){
int c = ;
int index = s.find('C');
string ans = "";
for(int i = index + ; i < s.size(); ++i) c = * c + s[i] - '';
while(c){
int k = c % ;
if(k == ){
ans += 'Z';
--c;
}
else ans += ('A' + k - );
c /= ;
}
reverse(ans.begin(), ans.end());
cout << ans << s.substr(, index-) << endl;
}
else{
int index = ;
string ans = "R";
int r = ;
for(; index < s.size(); ++index){
if('' <= s[index] && s[index] <= '') break;
r = * r + (s[index] - 'A' + );
}
ans += s.substr(index, s.size());
ans += 'C';
cout << ans << r << endl;
}
}
return ;
}
mdzz。。。
余弦定理求三角形内角，也是圆上的圆周角，乘以2得圆心角
求三圆心角最大公约数得正多边形每一份的圆心角ang，2*pi/ang得出边数n
海伦公式+正弦定理得三角形外接圆半径r = (a * b * c) / (4 * s)
s = n * r * r * sin(ang) / 2
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-;
double fgcd(double a, double b){
if (fabs(a) < eps)
return b;
if (fabs(b) < eps)
return a;
return fgcd(b, fmod(a, b));
}
double diameter(double a, double b, double c) {
double p = (a + b + c) / ;
double s = sqrt(p * (p - a) * (p - b) * (p - c));
return a * b * c / ( *s);
}
int main() {
double x[], y[], line[];
for(int i = ; i < ; ++i) scanf("%lf %lf", &x[i], &y[i]);
for(int i = ; i < ; ++i)
line[i] = sqrt((x[i] - x[(i+)%]) * (x[i] - x[(i+)%]) + (y[i] - y[(i+)%]) * (y[i] - y[(i+)%]));
double r = diameter(line[], line[], line[]);
double angle[];
for(int i = ; i < ; ++i)
angle[i] = acos( - line[i] * line[i] / ( * r * r));
angle[] = * acos(-) - angle[] - angle[];
double ang = fgcd(angle[], fgcd(angle[], angle[]));
printf("%.6f\n", r * r * sin(ang) / * ( * acos(-) / ang));
return ;
}
模拟水题 略麻烦(说到底是思路不清淅不严谨+弱)
#include <bits/stdc++.h>
using namespace std; struct info{
string name;
int score;
}aaa[];
map<string, int> game;
int main(){
int n;
cin >> n;
string name, ans;
int score;
for(int i = ; i < n; ++i){
cin >> name >> score;
aaa[i].name = name;
aaa[i].score = score;
game[name] += score;
}
int maxScore = -;
for(int i = ; i < n; ++i) maxScore = max(maxScore, game[aaa[i].name]);
map<string, int> tmp;
for(int i = ; i < n; ++i){
if(game[aaa[i].name] < maxScore) continue;
tmp[aaa[i].name] += aaa[i].score;
if(tmp[aaa[i].name] >= maxScore){
ans = aaa[i].name;
break;
}
}
cout << ans << endl;
return ;
}
然后就2B了.....
求2最少几个5最少几个然后min(num_two, num_five)就是答案 剩下输出路径
#include <bits/stdc++.h>
using namespace std; int matrix[][][];
int dp[][][];
int step[][][]; void print(int nx, int ny, int k){
if(nx == && ny == ) return ;
if(step[nx][ny][k]){
print(nx-, ny, k);
printf("D");
}
else{
print(nx, ny-, k);
printf("R");
}
}
int main(){
int n;
scanf("%d", &n);
int zerox = , zeroy = ;
for(int i = ; i <= n; ++i){
for(int j = ; j <= n; ++j){
int k;
scanf("%d", &k);
if(k == ){
zerox = i;
zeroy = j;
continue;
}
int numtwo = , numfive = ;
while(!(k & )){
++numtwo;
k >>= ;
}
while(k % == ){
++numfive;
k /= ;
}
matrix[i][j][] = numtwo;
matrix[i][j][] = numfive;
}
}
memset(dp, 0x3f, sizeof(dp));
dp[][][] = matrix[][][];
dp[][][] = matrix[][][];
for(int i = ; i <= n; ++i){
for(int j = ; j <= n; ++j){
if(i == && j == ) continue;
for(int k = ; k < ; ++k){
dp[i][j][k] = matrix[i][j][k] + min(dp[i-][j][k], dp[i][j-][k]);
if(dp[i-][j][k] < dp[i][j-][k]) step[i][j][k] = ;
}
}
}
int ans = min(dp[n][n][], dp[n][n][]);
if(ans == ) puts("");
else if(zerox && zeroy){
puts("");
int nx = , ny = ;
while(nx < zerox) ++nx, printf("D");
while(ny < zeroy) ++ny, printf("R");
while(nx < n) ++nx, printf("D");
while(ny < n) ++ny, printf("R");
return ;
}
else printf("%d\n", ans);
int k = dp[n][n][] < dp[n][n][] ? : ;
print(n, n, k);
return ;
}
模拟退火 先放着
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
#define _pow(a) ((a)*(a))
const double eps = 1e-; struct circle{ double x, y, r; }c[];
double angle[];
int cx[] = {, , , -}, cy[] = {, -, , }; double dis(double x1, double y1, double x2, double y2){
return sqrt(_pow(x1 - x2) + _pow(y1 - y2));
} double calc(double x, double y){
for(int i = ; i < ; ++i) angle[i] = dis(x, y, c[i].x, c[i].y) / c[i].r;
double value = ;
for(int i = ; i < ; ++i) value += _pow(angle[i] - angle[(i + ) % ]);
return value;
} int main(){
double x = , y = ;
for(int i = ; i < ; ++i){
scanf("%lf %lf %lf", &c[i].x, &c[i].y, &c[i].r);
x += c[i].x / 3.0;
y += c[i].y / 3.0;
}
double deviation = calc(x, y);
double step = ;
for(int i = ; i <= 1e5; ++i){
bool flag = false;
double xx, yy;
for(int j = ; j < ; ++j){
double nx = x + cx[j] * step, ny = y + cy[j] * step;
double nDeviation = calc(nx, ny);
if(nDeviation < deviation){
deviation = nDeviation;
flag = true;
xx = nx;
yy = ny;
}
}
if(flag){
x = xx;
y = yy;
}
else step /= 2.0;
}
if(deviation <= eps) printf("%.5f %.5f\n", x, y);
return ;
}
水题 但是大神的代码总是让我amazing，简单清晰直接
#include <bits/stdc++.h>
using namespace std; int main(){
char s[], t[];
scanf("%s %s", s, t);
int a = s[] - t[], b = s[] - t[];
char c = a > ? 'L' : (a = -a, 'R');
char d = b > ? 'D' : (b = -b, 'U');
printf("%d", a > b ? a : b);
while(a || b){
puts("");
if(a) --a, putchar(c);
if(b) --b, putchar(d);
}
return ;
}
分开按capacity排序 再维护前缀和 再枚举space为1的船的数量
代码好丑...........
#include <bits/stdc++.h>
using namespace std; struct vihecle{
int id;
int cpa;
}p1[], p2[];
int presum1[], presum2[]; bool cmp(vihecle a, vihecle b){
return a.cpa > b.cpa;
} int main(){
int n, v;
cin >> n >> v;
int i1 = , i2 = ;
presum1[] = presum2[] = ;
for(int i = ; i < n; ++i){
int t, pp;
cin >> t >> pp;
if(t == ){
p1[i1].cpa=pp;
p1[i1++].id=i+;
}
else{
p2[i2].cpa=pp;
p2[i2++].id=i+;
}
}
sort(p1, p1 + i1, cmp);
sort(p2, p2 + i2, cmp);
for(int i = ; i <= i1; ++i) presum1[i] = presum1[i-] + p1[i-].cpa;
for(int i = ; i <= i2; ++i) presum2[i] = presum2[i-] + p2[i-].cpa;
int ans = , ii1 = , ii2 = ;
for(int i = ; i <= i1; ++i){
if(i > v) break;
int tmp = presum1[i] + presum2[min((v-i)/, i2)];
if(tmp > ans){
ans = tmp;
ii1 = i;
ii2 = min((v - i) / , i2);
}
}
cout << ans << endl;
for(int i = ; i < ii1; ++i) cout << p1[i].id << " ";
for(int i = ; i < ii2; ++i) cout << p2[i].id << " ";
return ;
}
恶心的模拟..........讲真......玩起来容易写起来恶心.......
#include <bits/stdc++.h>
using namespace std; char g[][];
int c1, c2;
bool f1, f2; int main(){
for(int i = ; i < ; ++i) scanf("%s", g[i]);
c1 = c2 = ;
f1 = f2 = false; for(int i = ; i < ; ++i)
if(g[i][] == g[i][] && g[i][] == g[i][])
if(g[i][] == 'X') f1 = true;
else if(g[i][] == '') f2 = true;
for(int i = ; i < ; ++i)
if(g[][i] == g[][i] && g[][i] == g[][i])
if(g[][i] == 'X') f1 = true;
else if(g[][i] == '') f2 = true;
if(g[][] == g[][] && g[][] == g[][])
if(g[][] == 'X') f1 = true;
else if(g[][] == '') f2 = true;
if(g[][] == g[][] && g[][] == g[][])
if(g[][] == 'X') f1 = true;
else if(g[][] == '') f2 = true; for(int i = ; i < ; ++i)
for(int j = ; j < ; ++j)
if(g[i][j] == 'X') ++c1;
else if(g[i][j] == '') ++c2;
if(c1 == c2 || c1 == c2 + ){
if(f1 && f2) puts("illegal");
else if(f1 && !f2){
if(c1 == c2 + ) puts("the first player won");
else puts("illegal");
}
else if(!f1 && f2){
if(c1 == c2) puts("the second player won");
else puts("illegal");
}
else{
if(c1 + c2 == ) puts("draw");
else if(c1 + c2 < ){
if(c1 == c2) puts("first");
else if(c1 == c2 + ) puts("second");
}
else puts("illegal");
}
}
else puts("illegal");
return ;
}
对于处理中每一个过程都有'('不少于')'，将'?'置为')'，若'('少于')'则在前面的'?'中拿出转换价值最小的进行转换
#include <bits/stdc++.h>
using namespace std;
typedef long long ll; int main(){
string s;
cin >> s;
ll cnt = , ans = ;
priority_queue<pair<int, int> > q;
for(int i = ; i < s.size(); ++i){
if(s[i] == '(') ++cnt;
else if(s[i] == ')') --cnt;
else{
int a, b;
cin >> a >> b;
ans += b;
--cnt;
s[i] = ')';
q.push(make_pair(b - a, i));
}
if(cnt < ){
if(q.empty()) break;
pair<int, int> p = q.top();
q.pop();
ans -= p.first;
s[p.second] = '(';
cnt += ;
}
}
if(cnt) cout << "-1" << endl;
else cout << ans << endl << s << endl;
return ;
}
超级水......
#include <bits/stdc++.h>
using namespace std; int main(){
int n;
cin >> n;
cout << (((n & ) || n == ) ? "NO" : "YES") << endl;
return ;
}
水，随便贪一下
#include <bits/stdc++.h>
using namespace std; struct {
int mini, maxi;
}p[]; int main(){
int d, maxSum;
cin >> d >> maxSum;
int totalMin = , totalMax = ;
for(int i = ; i < d; ++i) cin >> p[i].mini >> p[i].maxi, totalMin += p[i].mini, totalMax += p[i].maxi;
if(totalMin > maxSum || totalMax < maxSum)
cout << "NO" << endl;
else{
cout << "YES" << endl;
int t = maxSum - totalMin;
for(int i = ; i < d; ++i){
int ansi = p[i].mini;
if(t){
ansi = min(p[i].maxi, ansi + t);
t -= (ansi - p[i].mini);
}
cout << ansi << " \n"[i == d-];
}
}
return ;
}
水，输入保证只有小写字母，map水过
#include <bits/stdc++.h>
using namespace std; int main(){
int n;
cin >> n;
map<string, int> name;
while(n--){
string s;
cin >> s;
if(name[s] == ) cout << "OK" << endl;
else cout << s << name[s] << endl;
++name[s];
}
return ;
}
n^2的暴力dp.......
#include <bits/stdc++.h>
using namespace std; int n;
struct env{
int w, h;
}p[];
int dp[], pre[]; int dfs(int k){
if(dp[k]) return dp[k];
for(int i = ; i <= n; ++i){
if(p[k].w < p[i].w && p[k].h < p[i].h){
if(dfs(i) + > dp[k]){
pre[k] = i;
dp[k] = dfs(i) + ;
}
}
}
return dp[k];
} int main(){
cin >> n;
for(int i = ; i <= n; ++i)
cin >> p[i].w >> p[i].h;
cout << dfs() << endl;
for(int i = pre[]; i > ; i = pre[i]) cout << i << " ";
cout << endl;
return ;
}
水
#include <bits/stdc++.h>
using namespace std; int main(){
string s;
int ans = , member = ;
while(getline(cin, s)){
if(s[] == '+') ++member;
else if(s[] == '-') --member;
else{
int len = s.end() - find(s.begin(), s.end(), ':') - ;
ans += member * len;
}
}
cout << ans << endl;
return ;
}
一眼样例明白大概但是奇数的情况要交替........
#include <bits/stdc++.h>
using namespace std; int main(){
vector<string> v;
string s;
int len = ;
while(getline(cin, s)){
v.push_back(s);
len = max(len, int(s.size()));
}
cout << string(len+, '*') << endl;
int k = -;
for(vector<string>::iterator i = v.begin(); i != v.end(); ++i){
int l = len - (*i).size();
string fr = string(floor(double(l)/), ' '), ba = string(ceil(double(l)/), ' ');
if(l & ){
if(!k) cout << "*" << ba << *i << fr << "*" << endl;
else cout << "*" << fr << *i << ba << "*" << endl;
k = ~k;
}
else cout << "*" << fr << *i << ba << "*" << endl;
}
cout << string(len+, '*') << endl;
return ;
}
dp，让我智障好久的题........
#include <bits/stdc++.h>
using namespace std; const int maxn = 1e6 + ;
int dp[maxn]; int main(){
ios::sync_with_stdio(false);
memset(dp, , sizeof(dp));
string s;
cin >> s;
stack<int> pos;
while(!pos.empty()) pos.pop();
int len = , sum = ;
for(int i = ; i < s.size(); ++i){
if(s[i] == '(') pos.push(i);
else if(!pos.empty()){
int j = pos.top();
pos.pop();
int tmpSize = i - j + ;
dp[i] = tmpSize + (j ? dp[j-] : );
if(dp[i] > len){
len = dp[i];
sum = ;
}
else if(dp[i] == len) ++sum;
}
}
cout << len << " " << sum << endl;
return ;
}
运动学卧槽......放弃
分类还是渣的不成样子.....
#include <bits/stdc++.h>
using namespace std; const double eps = 1e-;
double a, v, l, d, w, t1, t2, vd, vt, r; int fcmp(double x){
if(fabs(x) < eps) return ;
else return x < ? - : ;
} int main(){
scanf("%lf %lf %lf %lf %lf", &a, &v, &l, &d, &w);
vt = sqrt(w * w / + a * d);
r = l - d;
if(fcmp(d - v * v / / a) <= && fcmp(sqrt( * a * d) - w) <= ){
t1 = sqrt( * d / a);
vd = sqrt( * a * d);
}
else if(fcmp(d - v * v / / a) >= && fcmp(w - v) >= ){
t1 = d / v + v / / a;
vd = v;
}
else if(fcmp(v - vt) >= && fcmp(vt - w) >= ){
t1 = ( * vt - w) / a;
vd = w;
}
else{
t1 = d / v + ( * v - w) / a - ( * v * v - w * w) / / a / v;
vd = w;
}
if(fcmp(r - (v * v - vd * vd) / / a) >= )
t2 = r / v + (v - vd) / a - (v * v - vd * vd) / / a / v;
else
t2 = (sqrt( * a * r + vd * vd) - vd) / a;
printf("%.10f\n", t1 + t2);
return ;
}
找每个点左右分别严格高于这个点的点，有点像并查集的样子
还需要找处在中间的相同高度的点
每个点与左右两个高点及相同高度的点可以互相看到
#include <bits/stdc++.h>
using namespace std;
typedef long long ll; const int maxn = 1e6 + ;
int height[maxn], l[maxn], r[maxn], same[maxn];
ll ans = ; int main(){
int n;
scanf("%d", &n);
int maxi = -, pos = ;
for(int i = ; i < n; ++i){
scanf("%d", &height[i]);
if(height[i] > maxi){
maxi = height[i];
pos = i;
}
}
rotate(height, height + pos, height + n);
height[n] = maxi;
same[n] = ;
for(int i = n - ; i >= ; --i){
r[i] = i + ;
while(r[i] < n && height[i] > height[r[i]]) r[i] = r[r[i]];
if(r[i] != n && height[i] == height[r[i]]){
same[i] = same[r[i]] + ;
r[i] = r[r[i]];
}
}
for(int i = ; i <= n; ++i){
l[i] = i - ;
while(l[i] && height[i] >= height[l[i]]) l[i] = l[l[i]];
}
for(int i = ; i < n; ++i){
ans += ( + same[i]);
if(l[i] == && r[i] == n) --ans;
}
printf("%I64d\n", ans);
return ;
}
水题
#include <bits/stdc++.h>
using namespace std; int a, b, c, d; void init(){
int arr[];
cin >> arr[] >> arr[] >> arr[] >> arr[];
sort(arr, arr + );
a = arr[]; b= arr[]; c = arr[]; d = arr[];
}
int main(){
init();
if(a + b > c || a + b > d || a + c > d || b + c > d) puts("TRIANGLE");
else if(a + b == c || a + b == d || a + c == d || b + c == d) puts("SEGMENT");
else puts("IMPOSSIBLE");
return ;
}
两层广搜过的，然而智障了......用一层广搜+set就行了
两层的辣鸡代码....:
#include <bits/stdc++.h>
using namespace std; const int maxsize = ;
int cx[] = {, -, , }, cy[] = {, , , -};
struct pos{
int x, y;
pos(int xx = , int yy = ): x(xx), y(yy){};
}; char room[maxsize][maxsize];
bool flag[maxsize][maxsize]; int main(){
int n, m;
char color;
cin >> n >> m;
cin.get();
cin >> color;
queue<pos> mainpos;
memset(flag, false, sizeof(flag));
for(int i = ; i < n; ++i)
for(int j = ; j < m; ++j){
cin >> room[i][j];
if(room[i][j] == color){
mainpos.push(pos(i, j));
flag[i][j] = true;
}
}
int ans = ;
while(!mainpos.empty()){
int x = mainpos.front().x, y = mainpos.front().y;
mainpos.pop();
for(int i = ; i < ; ++i){
int nx = x + cx[i], ny = y + cy[i];
if(nx < || n <= nx || ny < || m <= ny || flag[nx][ny] || room[nx][ny] == '.') continue;
++ans;
queue<pos> tpos;
tpos.push(pos(nx, ny));
flag[nx][ny] = true;
char tcolor = room[nx][ny];
while(!tpos.empty()){
int xx = tpos.front().x, yy = tpos.front().y;
tpos.pop();
for(int j = ; j < ; ++j){
int nxx = xx + cx[j], nyy = yy + cy[j];
if(nxx < || n <= nxx || nyy < || m <= nyy || room[nxx][nyy] != tcolor || flag[nxx][nyy]) continue;
flag[nxx][nyy] = true;
tpos.push(pos(nxx, nyy));
}
}
}
}
cout << ans << endl;
return ;
}
一层 + set：
#include <bits/stdc++.h>
using namespace std; const int maxsize = ;
int cx[] = {, -, , }, cy[] = {, , , -};
struct pos{
int x, y;
pos(int xx = , int yy = ): x(xx), y(yy){};
}; char room[maxsize][maxsize]; int main(){
int n, m;
char color;
cin >> n >> m;
cin.get();
cin >> color;
queue<pos> mainpos;
for(int i = ; i < n; ++i)
for(int j = ; j < m; ++j){
cin >> room[i][j];
if(room[i][j] == color)
mainpos.push(pos(i, j));
}
set<char> ans;
while(!mainpos.empty()){
int x = mainpos.front().x, y = mainpos.front().y;
mainpos.pop();
for(int i = ; i < ; ++i){
int nx = x + cx[i], ny = y + cy[i];
if(nx < || n <= nx || ny < || m <= ny || room[nx][ny] == color || room[nx][ny] == '.') continue;
ans.insert(room[nx][ny]);
}
}
cout << ans.size() << endl;
return ;
}
水题，第一次用deque
#include <bits/stdc++.h>
using namespace std; int main(){
int n;
cin >> n;
deque<int> chb;
for(int i = ; i < n; ++i){
int tmp;
cin >> tmp;
chb.push_back(tmp);
}
int t1 = chb.front(), ans1 = , t2 = , ans2 = ;
chb.pop_front();
while(!chb.empty()){
if(t1 <= t2){
t1 += chb.front();
++ans1;
chb.pop_front();
}
else{
t2 += chb.back();
++ans2;
chb.pop_back();
}
}
cout << ans1 << " " << ans2 << endl;
return ;
}
不会做.....题解看不懂....... dfs
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = ;
int n, a, b, ans, h[MAXN], hh[MAXN]; bool dfs(int curpos, int time, int hitpos){
if(time == ){
for(int i = ; i <= n; ++i)
if(h[i] >= ) return false;
return true;
}
if(h[curpos] < ) return dfs(curpos+, time, hitpos);
for(int i = min(n-, max(hitpos, max(, curpos))); i <= min(n-, curpos+); ++i){
h[i] -= a; h[i-] -= b; h[i+] -= b;
if(dfs(curpos, time-, i)){
hh[time] = i;
return true;
}
h[i] += a; h[i-] += b; h[i+] += b;
}
return false;
} int main(){
cin >> n >> a >> b;
for(int i = ; i <= n; ++i) cin >> h[i];
for(int i = ; ; ++i){
memset(hh, , sizeof(hh));
if(dfs(, i, )){
cout << i << endl;
for(int j = ; j <= i; ++j)
cout << hh[j] << " ";
return ;
}
}
return ;
}
求最大值最小值差不超过给定值的最长子串
ST表预处理区间最大值最小值，加个二分右边界...
#include <bits/stdc++.h>
using namespace std; const int maxn = 1e5 + ;
int n, k;
int data[maxn];
int premin[maxn][], premax[maxn][];
struct tpoint{
int l, r;
tpoint(int _l, int _r): l(_l), r(_r) {};
}; void init_st(){
for(int i = ; i <= n; ++i)
premin[i][] = premax[i][] = data[i];
int k = floor(log(double(n)) / log(2.0));
for(int j = ; j <= k; ++j)
for(int i = n; i >= ; --i)
if(i + (<<(j-)) <= n){
premin[i][j] = min(premin[i][j-], premin[i+(<<(j-))][j-]);
premax[i][j] = max(premax[i][j-], premax[i+(<<(j-))][j-]);
}
} int query_st(int l, int r){
int k = floor(log(double(r-l+)) / log(2.0));
return max(premax[l][k], premax[r-(<<k)+][k]) - min(premin[l][k], premin[r-(<<k)+][k]);
} int check(int i){
int l = i, r = n + ;
while(l < r - ){
int mid = (l + r) >> ;
if(query_st(i, mid) <= k) l = mid;
else r = mid;
}
return l;
} int main(){
scanf("%d %d", &n, &k);
for(int i = ; i <= n; ++i)
scanf("%d", &data[i]);
init_st();
int amount = ;
vector<tpoint> vp;
for(int l = ; l <= n; ++l){
int r = check(l);
int num = r - l + ;
if(num > amount){
vp.clear();
vp.push_back(tpoint(l, r));
amount = num;
}
else if(num == amount) vp.push_back(tpoint(l, r));
}
printf("%d %d\n", amount, vp.size());
for(vector<tpoint>::iterator i = vp.begin(); i != vp.end(); ++i)
printf("%d %d\n", (*i).l, (*i).r);
return ;
}
水题但是有点意思(因为你还是个渣啊~)
记下每一行每一列多少个黑，凑齐8个答案+1，若答案是16就-8
#include <bits/stdc++.h>
using namespace std; char cb[][];
int cnt[][]; int main(){
ios::sync_with_stdio(false);
int ans = ;
for(int i = ; i < ; ++i)
for(int j = ; j < ; ++j){
cin >> cb[i][j];
if(cb[i][j] == 'B'){
++cnt[i][];
++cnt[][j];
}
if(cnt[i][] == ) ++ans;
if(cnt[][j] == ) ++ans;
}
if(ans == ) ans -= ;
cout << ans << endl;
return ;
}
粗暴地模拟，大概坑都是给我挖的....
#include <bits/stdc++.h>
using namespace std; const int maxn = ;
int n, k, mem[maxn]; int main(){
int counter = ;
scanf("%d %d", &n, &k);
while(n--){
char s[];
int x;
scanf("%s", s);
if(s[] == 'a' || s[] == 'e'){
scanf("%d", &x);
if(s[] == 'a'){
bool flag = false;
int tmp = ;
for(int i = ; i < k; ++i){
if(mem[i] == ) ++tmp;
else tmp = ;
if(tmp == x){
printf("%d\n", ++counter);
int index = i;
while(tmp--) mem[index--] = counter;
flag = true;
break;
}
}
if(!flag) puts("NULL");
}
else{
bool flag = false;
for(int i = ; i < k; ++i){
if(x == ) break;
if(mem[i] == x){
mem[i] = ;
flag = true;
}
}
if(!flag) puts("ILLEGAL_ERASE_ARGUMENT");
}
}
else{
for(int i = ; i < k; ++i){
if(mem[i] != ){
int j = i;
while(j > && mem[j-] == ) --j;
swap(mem[i], mem[j]);
}
}
}
}
return ;
}
扩展欧几里得解不定方程组
#include <bits/stdc++.h>
using namespace std;
typedef long long ll; ll ex_gcd(ll a, ll b, ll& x, ll& y){
if(b == ){
x = ;
y = ;
return a;
}
int k = ex_gcd(b, a % b, x, y);
int t = y;
y = x - (a / b) * y;
x = t;
return k;
} int main(){
ios::sync_with_stdio(false);
ll a, b, c, x, y;
cin >> a >> b >> c;
ll k = ex_gcd(a, b, x, y);
c = -c;
if(c % k) cout << "-1" << endl;
else cout << (c / k * x) << " " << (c / k * y) << endl;
return ;
}
字符串两个方向hash判回文+dp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll; const int maxn = 5e6 + ;
const int mod = 1e9 + ;
const int tmp = ;
char s[maxn];
ll prefix[maxn], suffix[maxn], dp[maxn]; int main(){
gets(s);
for(int i = ; s[i]; ++i){
if(s[i] >= '' && s[i] <= '')
s[i] -= '';
else if(s[i] >= 'a' && s[i] <= 'z')
s[i] -= ('a' - );
else
s[i] -= ('A' - );
}
ll len = strlen(s), base = ;
for(int i = ; i <= len; ++i){
prefix[i] = (prefix[i-] + s[i-] * base) % mod;
base = (base * tmp) % mod;
suffix[i] = (suffix[i-] * tmp + s[i-]) % mod;
}
ll ans = ;
for(int i = ; i <= len; ++i){
if(prefix[i] == suffix[i]){
dp[i] = dp[i>>] + ;
ans += dp[i];
}
}
printf("%I64d\n", ans);
return ;
}
arhgoiawhengiuehgoaijgioj........
水题，用py3写是因为.....懒......
s = input()
s1 = input()
s2 = input() forw = backw = False
tmp = s.find(s1)
if ~tmp:
if ~s[tmp+len(s1):].find(s2):
forw = True
s = s[::-1]
tmp = s.find(s1)
if ~tmp:
if ~s[tmp+len(s1):].find(s2):
backw = True if forw and backw: print('both')
elif forw: print('forward')
elif backw: print('backward')
else: print('fantasy')
啊啊啊啊啊啊啊手贱啊啊啊啊啊啊啊 真·教育题 谢谢(微笑)
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> point;
map<point, int> check;
int cx[] = {, -, , }, cy[] = {, , -, };
int main(){
string s;
cin >> s;
int x = , y = , xx = , yy = ;
++check[point(x, y)];
bool flag = true;
for(int i = ; i < s.size(); ++i){
if(s[i] == 'L') --x;
else if(s[i] == 'R') ++x;
else if(s[i] == 'U') ++y;
else --y;
for(int i = ; i < ; ++i){
int xxx = x + cx[i], yyy = y + cy[i];
if(xxx == xx && yyy == yy) continue;
if(check[point(xxx, yyy)]){
flag = false;
break;
}
}
if(check[point(x, y)]) flag = false;
if(!flag) break;
++check[point(x, y)];
xx = x; yy = y;
}
cout << (flag ? "OK" : "BUG") << endl;
return ;
}
状压dp，不需要绝对固定的顺序，所以每次只要取最前面可取的(line34的break)，每次拿一个或两个(最里层for从 j 开始)
#include <bits/stdc++.h>
using namespace std;
const int MAXN = ;
const int INF = 0x3f3f3f3f;
struct point{
int x, y;
point(){}
point(int xx, int yy): x(xx), y(yy) {}
};
point p[MAXN];
int n, dp[<<MAXN], pre[<<MAXN], dis[MAXN][MAXN], ans[];
inline int calc(point a, point b){
return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
int main(){
cin >> p[].x >> p[].y >> n, p[n] = p[];
for(int i = ; i < n; ++i) cin >> p[i].x >> p[i].y;
for(int i = ; i <= n; ++i) for(int j = ; j <= n; ++j) dis[i][j] = calc(p[i], p[j]);
memset(dp, INF, sizeof(dp));
dp[] = ;
for(int i = ; i < (<<n); ++i)
if(dp[i] != INF)
for(int j = ; j < n; ++j)
if(!(i>>j&)){
for(int k = j; k < n; ++k)
if(!(i>>k&)){
int cur = i | (<<j) | (<<k);
int tmp = dp[i] + dis[n][j] + dis[j][k] + dis[k][n];
if(dp[cur] > tmp){
dp[cur] = tmp;
pre[cur] = i;
}
}
break;
}
cout << dp[(<<n)-] << endl << ;
int cnt = ;
for(int i = (<<n)-; i; i = pre[i]){
int tmp = pre[i] ^ i;
ans[cnt++] = ;
for(int j = ; j < n; ++j)
if((<<j)&tmp)
ans[cnt++] = j + ;
}
for(int i = cnt-; ~i; --i) cout << " " << ans[i];
return ;
}
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