codeforces 1136A:

题意:一本书有n个章节,每个章节的分别在li到ri页,小明读完书后将书折在第k页,问还有多少章节没有读

题解:控制k在li~ri的范围内后输出n-i即可

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
LL read() {
int x = , f = ; char ch = getchar();
while(ch < '' || ch > '') {
if(ch == '-')f = -;
ch = getchar();
}
while(ch >= '' && ch <= '') {
x = x * + ch - '';
ch = getchar();
}
return x * f;
}
const double eps = 1e-;
const int mod = 1e9 + ;
const int maxn = 2e5 + ;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;
int l[maxn];
int r[maxn];
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int n, k;
scanf("%d", &n);
for(int i = ; i < n; i++) {
scanf("%d%d", &l[i], &r[i]);
}
scanf("%d", &k);
for(int i = ; i < n; i++) {
if(l[i] <= k && k <= r[i]) {
cout << n - i << endl;
return ;
}
}
return ;
}

codeforces 1136B:
题意:有n个井盖,井盖上有一颗石头,你每次可以做以下三种操作中的一种

1.将石头丢在隔壁的井盖上

2.走向相邻的井盖

3.如果井盖上没有石头,将井盖打开

你初始时在第k个井盖上,问你最少需要多少次操作可以打开全部的井盖

题解:每个井盖都需要进行3次操作才能到这个井盖上,首先我们需要到这个井盖上,然后我们需要搬石头,再然后我们要打开井盖,但是我们会走重复的路径,所以要想操作数最少,我们应该走重复的路最少,即min(n-k,k-1)

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
LL read() {
int x = , f = ; char ch = getchar();
while(ch < '' || ch > '') {
if(ch == '-')f = -;
ch = getchar();
}
while(ch >= '' && ch <= '') {
x = x * + ch - '';
ch = getchar();
}
return x * f;
}
const double eps = 1e-;
const int mod = 1e9 + ;
const int maxn = 2e5 + ;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f; int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int n,k;
scanf("%d%d", &n, &k);
cout << min(k - , n - k) + * n << endl;
return ;
}

codeforces 1136C:

题意:给你一个矩阵A和一个矩阵B,问你矩阵A和矩阵B是否是同类矩阵(转置矩阵也是同类的

题解:记录一下两个矩阵对角线上的元素是否相等即可

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
LL read() {
int x = , f = ; char ch = getchar();
while(ch < '' || ch > '') {
if(ch == '-')f = -;
ch = getchar();
}
while(ch >= '' && ch <= '') {
x = x * + ch - '';
ch = getchar();
}
return x * f;
}
const double eps = 1e-;
const int mod = 1e9 + ;
const int maxn = 2e5 + ;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;
int a[][];
int b[][];
vector<int> vec1[], vec2[];
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int n, m;
scanf("%d%d", &n, &m);
for(int i = ; i < n; i++) {
for(int j = ; j < m; j++) {
scanf("%d", &a[i][j]);
}
}
for(int i = ; i < n; i++) {
for(int j = ; j < m; j++) {
scanf("%d", &b[i][j]);
}
}
for(int i = ; i < n; i++) {
for(int j = ; j < m; j++) {
vec1[i + j].push_back(a[i][j]);
vec2[i + j].push_back(b[i][j]);
}
}
for(int i = ; i < n + m - ; i++) {
sort(vec1[i].begin(), vec1[i].end());
sort(vec2[i].begin(), vec2[i].end());
if(vec1[i] != vec2[i]) {
cout << "NO" << endl;
return ;
}
}
cout << "YES" << endl;
return ;
}

codeforces 1136D:

题意:有一队人,其中有n对人可以两两交换,你在队伍的尾部,问你你最多可以向前移动多少步

题解:用一个vector记录你可以交换的人,然后判断你是否能与vector里面的人进行交换,如果你可以与vector里面所有的人交换的话,你就可以移动一步

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
LL read() {
int x = , f = ; char ch = getchar();
while(ch < '' || ch > '') {
if(ch == '-')f = -;
ch = getchar();
}
while(ch >= '' && ch <= '') {
x = x * + ch - '';
ch = getchar();
}
return x * f;
}
const double eps = 1e-;
const int mod = 1e9 + ;
const int maxn = 2e6 + ;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;
set<pair<int, int> > s;
int a[maxn];
vector<int> vec;
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int n, m;
scanf("%d%d", &n, &m);
for(int i = ; i <= n; i++) {
scanf("%d", &a[i]);
}
vec.push_back(a[n]);
for(int i = , u, v; i < m; i++) {
scanf("%d%d", &u, &v);
s.insert(make_pair(u, v));
}
int ans = ;
for(int i = n - ; i >= ; i--) {
int flag = ;
for(int j = ; j < vec.size(); j++) {
if(s.count(make_pair(a[i], vec[j])) == ) {
flag = ;
break;
}
}
if(flag) {
ans++;
} else {
vec.push_back(a[i]);
}
} cout << ans << endl;
return ;
}

codeforces 1136E:

题意:给你一个长度为n的序列a和长度为n-1的序列k,序列a在任何时候都满足如下性质,a[i+1]>=ai+ki,如果更新后a[i+1]<ai+ki了,那么a[i+1]=ai+ki

现在给你q次操作

操作1:将位置为pos的元素+x

操作2:询问区间l,r的区间和

题解:非常明显的线段树题,我们不好维护的是,如果更新后,当前数字变大到不满足限制条件时,我后面的元素也要做出相应的更新

那么我们就将a序列先减去k序列,这样的a序列也是满足限制条件了,然后我们记录下k的前缀和的前缀和,避免询问时缺少k的贡献,

数学推导如下

序列a满足单调不减性,

则∑ai 同样满足单调不减性,

当我们对位置为pos的元素进行更新时,

如果后面的元素 a[pos+R]<a[pos],则该元素要被覆盖,

所以我们二分右端点,将区间【pos,R】覆盖为a[pos]+x即可

为了避免重复计算ki对a的贡献所以我们覆盖区间时可以用如下技巧

用c来记录k的前缀和的前缀和

每次覆盖时,我们将区间【l,r】覆盖为(a[pos]+x-k[pos])*(r-l+1)+c[r+1]-c[l]

这样就不会使得k的贡献计算错了

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
LL read() {
int x = , f = ; char ch = getchar();
while(ch < '' || ch > '') {
if(ch == '-')f = -;
ch = getchar();
}
while(ch >= '' && ch <= '') {
x = x * + ch - '';
ch = getchar();
}
return x * f;
}
const int maxn = 2e5 + ;
LL sum[maxn << ];
LL lazy[maxn << ];
LL a[maxn];
LL k[maxn];
LL sub[maxn];
LL c[maxn];
LL NUL = -1e18;
void push_up(int rt) {
sum[rt] = (LL)sum[rt << ] + sum[rt << | ];
}
void push_down(int l, int r, int rt) {
if(lazy[rt] == NUL) return;
int mid = (l + r) >> ;
lazy[rt << ] = lazy[rt];
lazy[rt << | ] = lazy[rt];
sum[rt << ] = (LL)(mid - l + ) * lazy[rt] + c[mid + ] - c[l];
sum[rt << | ] = (LL)(r - mid) * lazy[rt] + c[r + ] - c[mid + ];
lazy[rt] = NUL;
}
void build(int l, int r, int rt) {
lazy[rt] = NUL;
if(l == r) {
sum[rt] = a[l];
return;
}
int mid = (l + r) >> ;
build(l, mid, rt << );
build(mid + , r, rt << | );
push_up(rt);
}
void update(int L, int R, LL val, int l, int r, int rt) {
if(L <= l && r <= R) {
sum[rt] = (LL)(r - l + ) * val + c[r + ] - c[l];
lazy[rt] = val;
return;
}
int mid = (l + r) >> ;
push_down(l, r, rt);
if(L <= mid) update(L, R, val, l, mid, rt << );
if(R > mid) update(L, R, val, mid + , r, rt << | );
push_up(rt);
}
LL query(int L, int R, int l, int r, int rt) {
if(L <= l && r <= R) {
//debug2(rt,sum[rt]);
return sum[rt];
}
int mid = (l + r) >> ;
push_down(l, r, rt);
LL ans = ;
if(L <= mid) ans += query(L, R, l, mid, rt << );
if(R > mid) ans += query(L, R, mid + , r, rt << | );
return ans;
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int n;
scanf("%d", &n);
for(int i = ; i < n; i++) {
scanf("%lld", &a[i]);
}
for(int i = ; i < n; i++) {
scanf("%lld", &k[i]);
k[i] = k[i] + k[i - ];
}
for(int i = ; i <= n; i++) {
c[i] = c[i - ] + k[i - ];
}
build(, n - , );
int q;
cin >> q; while(q--) {
char op[];
int l, r, pos;
LL val;
cin >> op; if(op[] == '+') {
cin >> pos >> val;
pos--;
int l = pos;
int r = n - ;
LL tmp = query(pos, pos, , n - , );
while(l < r) {
int mid = (l + r + ) / ;
if(tmp + val + k[mid] - k[pos] > query(mid, mid, , n - , )) {
l = mid;
} else {
r = mid - ;
}
}
// debug3(pos, r, tmp + val - k[pos]);
update(pos, r, tmp + val - k[pos], , n - , );
} else {
cin >> l >> r;
cout << query(l - , r - , , n - , ) << endl;;
// printf("%lld\n", query(l - 1, r - 1, 0, n - 1, 1));
}
}
}

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