Semana i 2018

A Giga-Kilo-Gigabyte

思路: dp水题

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);

//head

const int MOD = 1e9 + , N = 1e6 + ;

LL dp[N];

int main() {

    int T, n;

    dp[] = ;

    for (int i = ; i < N; i++) {

        for (int j = max(, i-); j < i; j++) dp[i] = (dp[i] + dp[j]) % MOD;

    }

    scanf("%d", &T);

    for (int i = ; i <= T; i++) {

        scanf("%d", &n);

        printf("%lld\n", dp[n/]);

    }

    return ;

}

B Three Couse Meal

思路:用set

n^2logn暴力枚举

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);

//head

const int N = 2e3 + ;

pair<int, pii> a[N];

multiset<pii> s1;

multiset<int> s2;

int main() {

    int n, k;

    scanf("%d %d", &n, &k);

    for (int i = ; i <= n; i++) scanf("%d %d %d", &a[i].fi, &a[i].se.fi, &a[i].se.se);

    sort(a+, a++n);

    LL ans = 1e16;

    for (int i = ; i <= n; i++) {

        s1.insert(a[i].se);

        if(i >= k) {

            s2.clear();

            for (auto it : s1) {

                s2.insert(it.se);

                if(s2.size() > k) s2.erase(--s2.end());

                if(s2.size() == k) {

                    ans = min(ans, 1LL*a[i].fi + 1LL*it.fi + 1LL*(*--s2.end()));

                }

            }

        }

    }

    printf("%lld\n", ans);

    return ;

}

C Basic Encryption

思路:水题, 减S取模

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);

//head

const int N = 1e3 + ;

char s[N];

bool isAlphaBete(char c){

    return c>='A'&&c<='Z'||c>='a'&&c<='z';

}

int main() {

    int n, S;

    scanf("%d %d", &n, &S);

    S %= ;

    char c;

    c = getchar();

    while((c = getchar())!=EOF){

        if(isupper(c)) {

            int t = c - 'A';

            t = (t +  - S) % ;

            c = 'A' + t;

        }

        else if(islower(c)) {

            int t = c - 'a';

            t = (t +  - S) % ;

            c = 'a' + t;

        }

        putchar(c);

    }

    return ;

}

D Freddy and minifier

思路:贪心

假设相邻两个元素a, b, 在此之前的大象的体重为w

那么按a前b后的顺序的花费是 w*a.c + w*a.r*b.c

按b前a后的顺序的花费是 w*b.c + w*b.r*a.c

无论什么顺序,经过这两个元素后大象的体重都是w*a.r*b.r

那么a在b前的条件是w*a.c + w*a.r*b.c < w*b.c + w*b.r*a.c

即 a.c + a.r*b.c < b.c + b.r*a.c

按这个条件写个cmp函数排个序即可

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);

//head

const int N = 1e5 + ;

const LL INF = (1ll<<);

struct node {

    int c;

    double r;

}a[N];

int n, w;

bool cmp(node a, node b) {

    return  a.c + b.c*a.r < b.c + a.c*b.r;

}

void solve() {

    double ans = INF;

    sort(a+, a++n, cmp);

    double tw = w, tans = ;

    for (int i = ; i <= n; i++) {

        tans += tw * a[i].c;

        tw *= a[i].r;

    }

    ans = min(ans, tans);

    printf("%.10f\n", ans);

}

int main() {

    scanf("%d %d", &n, &w);

    for (int i = ; i <= n; i++) scanf("%d %lf", &a[i].c, &a[i].r);

    solve();

    return ;

}

E Perfect Polygon

思路:一个向量旋转乘以下面的旋转矩阵即可(Θ > 0 表示逆时针)

然后多边形旋转的话就一个点不动,其他点绕改点旋转

旋转好后将多边形左下角移动到与x轴正半轴和y正半轴相切,然后按w*h等比例缩放

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);

//head

const int N = 1e5 + ;

const int INF = 0x7f7f7f7f;

const double eps = 1e-;

//考虑误差的加法

double add(double a, double b) {

    if(fabs(a + b) < eps * (fabs(a) + fabs(b))) return ;

    return a + b;

}

//考虑误差的与0比较

int dcmp(double x) {

    if(fabs(x) < eps) return ;

    else return x<?-:;

}

struct P {

    double x, y;

    P(){}

    P(double x, double y) :x(x), y(y){}

    bool operator == (P p) {

        return dcmp(x - p.x) ==  && dcmp(y - p.y) == ;

    }

    P operator + (P p) {

        return P(add(x, p.x), add(y, p.y));

    }

    P operator - (P p) {

        return P(add(x, -p.x), add(y, -p.y));

    }

    P operator * (double p) {

        return P(x * p, y * p);

    }

    P operator / (double p) {

        return P(x / p, y / p);

    }

    //点积

    double dot(P p) {

        return add(x * p.x, y * p.y);

    }

    //叉积

    double cross(P p) {

        return add(x * p.y, -y * p.x);

    }

}p[N];

typedef P Vector;

//向量逆时针旋转

Vector Rotate(Vector a,double rad)  {

    return Vector(a.x * cos(rad) - a.y * sin(rad), a.x * sin(rad) + a.y * cos(rad));

}

int main() {

    int a, w, h, n;

    scanf("%d %d %d %d", &a, &w, &h, &n);

    double A = a*pi/;

    for (int i = ; i <= n; i++) scanf("%lf %lf", &p[i].x, &p[i].y);

    for (int i = ; i <= n; i++) p[i] = p[] + Rotate(p[i] - p[], A);

    double mnx = INF, mny = INF;

    for (int i = ; i <= n; i++) mnx = min(mnx, p[i].x), mny = min(mny, p[i].y);

    for (int i = ; i <= n; i++) p[i].x -= mnx, p[i].y -= mny;

    double mxx = , mxy = ;

    for (int i = ; i <= n; i++) mxx = max(mxx, p[i].x), mxy = max(mxy, p[i].y);

    for (int i = ; i <= n; i++) p[i].x = p[i].x*w/mxx, p[i].y = p[i].y*h/mxy;

    for (int i = ; i <= n; i++) printf("%.10f %.10f\n", p[i].x, p[i].y);

    return ;

}

F Minimum Played Times

思路:首先将给的小数乘以10000得到式子 d / 100000 中的 d, 然后10000/gcd(10000, d) 就是答案

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);

//head

char s[];

int main() {

    int T;

    scanf("%d", &T);

    for (int i = ; i <= T; i++) {

        scanf("%s", s);

        int n = strlen(s);

        int t = ;

        for (int i = ; i < n; i++) {

            t = t* + s[i] - '';

        }

        for (int i = n; i < ; i++) {

            t = t*;

        }

        int d = __gcd(t, );

        printf("%d\n", /d);

    }

    return ;

}

G A+B+C

思路:水题, (a*d*f + c*b*f + e*b*d) / (b*d*f) 约分后即是答案

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);

//head

int main() {

    int T;

    LL a, b, c, d, e, f;

    scanf("%d", &T);

    for (int i = ; i <= T; i++) {

        scanf("%lld/%lld %lld/%lld %lld/%lld", &a, &b, &c, &d, &e, &f);

        LL up = a*d*f + c*b*f + e*b*d;

        LL down = b*d*f;

        LL d = __gcd(up, down);

        up /= d;

        down /= d;

        printf("%lld/%lld\n", up, down);

    }

    return ;

}

H Diego and drinks

思路:水题, 题意难懂, 把a drink理解为一瓶酒

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);

//head

const int MOD = 1e9 + , N = 1e6 + ;

LL dp[N];

int main() {

    int T, n, k;

    scanf("%d", &T);

    for (int i = ; i <= T; i++) {

        scanf("%d %d", &n, &k);

        int b = n*(n+)/;

        if(k < b) puts("Too drunk to count");

        else if((k-b) %  n == ) printf("%d\n", n++(k-b)/n);

        else puts("Too drunk to count");

    }

    return ;

}

I Water in BeagleTown

思路: 最大流, 要拆点

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);

//head

const int N = 2e3 + , M = 6e3 + ;

const int INF = 0x3f3f3f3f;

pii a[N];

piii b[N];

struct edge {

    int to, cap, rev;

};

vector<edge> g[M];

int level[M], iter[M], mxv = ;

void add_edge(int from, int to, int cap) {

   g[from].pb({to, cap, g[to].size()});

   g[to].pb({from, , g[from].size()-});

}

void bfs(int s) {

    for (int i = ; i <= mxv; i++) level[i] = -;

    queue<int> q;

    level[s] = ;

    q.push(s);

    while(!q.empty()) {

        int v = q.front();

        q.pop();

        for (edge &e : g[v]) {

            if(e.cap >  && level[e.to] < ) {

                level[e.to] = level[v] + ;

                q.push(e.to);

            }

        }

    }

}

int dfs(int v, int t, int f) {

    if (v == t) return f;

    for (int &i = iter[v]; i < g[v].size(); i++) {

        edge &e = g[v][i];

        if(e.cap >  && level[v] < level[e.to]) {

            int d = dfs(e.to, t, min(f, e.cap));

            if(d > ) {

                e.cap -= d;

                g[e.to][e.rev].cap += d;

                return d;

            }

        }

    }

    return ;

}

int max_flow(int s, int t) {

    int flow = ;

    for(;;) {

        bfs(s);

        if(level[t] < ) return flow;

        for (int i = ; i <= mxv; i++) iter[i] = ;

        int f;

        while((f = dfs(s, t, INF)) > ) {

            flow += f;

        }

    }

}

int main() {

    int n, m;

    LL s;

    scanf("%d %d %lld", &n, &m, &s);

    if(s < ) return *puts("NO");

    s -= ;

    for (int i = ; i <= n; i++) scanf("%d %d", &a[i].fi, &a[i].se);

    for (int i = ; i <= m; i++) scanf("%d %d %d", &b[i].fi.fi, &b[i].fi.se, &b[i].se);

    int st = ;

    for (int i = ; i <= n; i++) add_edge(st, i, );

    for (int i = ; i <= n; i++) {

        for (int j = ; j <= m; j++) {

            int dx = abs(b[j].fi.fi - a[i].fi), dy = abs(b[j].fi.se - a[i].se);

            if(1ULL*dx*dx + 1ULL*dy*dy <= 1ULL*s*s) {

                add_edge(i, n+j, );

            }

        }

    }

    for (int i = ; i <= m; i++) {

        add_edge(i+n, i+n+m, b[i].se);

    }

    mxv = m+n+m+;

    for (int i = ; i <= m; i++) {

        add_edge(i+n+m, mxv, INF);

    }

    if(max_flow(st, mxv) == n) puts("YES");

    else puts("NO");

    return ;

}

J Luca and Stock

思路:线段树, 用long double

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);

//head

const int N = 1e5 + ;

const long double up = pow((long double)2.0, );

long double tree[N<<];

void push_up(int rt) {

    tree[rt] = tree[rt<<] * tree[rt<<|];

}

void build(int rt, int l, int r) {

    if(l == r) {

        cin >> tree[rt];

        return ;

    }

    int m = l+r >> ;

    build(ls);

    build(rs);

    push_up(rt);

}

void update(int p, double x, int rt, int l, int r) {

    if(l == r) {

        tree[rt] = (long double)x;

        return ;

    }

    int m = l+r >> ;

    if(p <= m) update(p, x, ls);

    else update(p, x, rs);

    push_up(rt);

}

long double query(int L, int R, int rt, int l, int r) {

    if(L <= l && r <= R) return tree[rt];

    int m = l+r >> ;

    long double ans = ;

    if(L <= m) ans *= query(L, R, ls);

    if(R > m) ans *= query(L, R, rs);

    return ans;

}

int main() {

    int n;

    scanf("%d", &n);

    build(, , n);

    int q, ty, l, r, x;

    double p;

    scanf("%d", &q);

    for (int i = ; i <= q; i++) {

        scanf("%d", &ty);

        if(ty == ) {

            scanf("%d %lf", &x, &p);

            update(x, p, , , n);

        }

        else {

            scanf("%d %d", &l, &r);

            long double t = query(l, r, , , n);

            if(t >= up) printf("INFINITE!\n");

            else cout << fixed << setprecision() << t << '\n';

        }

    }

    return ;

}

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