HZNU 2019 Summer training 8

A - Petya and Origami

 CodeForces - 1080A 

题意：制造一份邀请函需要2份a物品，5份b物品，8份c物品，一个盒子里面有k份物品（可以为a或b或c）问你制造n份邀请函需要用多少个盒子

题解：加起来就行了

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include<stack>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;const int maxn = 4e4 + 10;
const int mod = 1e9 + 7;int main()
{int n,k;cin >> n >> k;int sum = 0;sum += ceil(2.0 * n / (k * 1.0));sum += ceil(5.0 * n / (k * 1.0));sum += ceil(8.0 * n / (k * 1.0));cout << sum << endl;
}

 

B - Margarite and the best present

 CodeForces - 1080B 

题意：区间内偶数和减去奇数和

题解：分类一下就好了

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include<stack>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;const int maxn = 4e4 + 10;
const int mod = 1e9 + 7;int main()
{int t;cin >> t;while (t--){ll l, r;cin >> l >> r;ll ans;if (l == r){if (l % 2 == 0)ans = l;elseans = -1 * l;}else{if (l % 2 == 1 && r % 2 == 1)ans = (r - l) / 2 - r;else if (l % 2 == 1 && r % 2 == 0)ans = (r - l + 1) / 2;else if (l % 2 == 0 && r % 2 == 0)ans = -1*(r - l) / 2 + r;elseans = ((r - l + 1) / 2) * (-1);}cout << ans << endl;}}

 

C - Masha and two friends

 CodeForces - 1080C 

题意：给你一个n行m列的黑白块相间的棋盘，进行两次操作，第一次把（x1,y1）到（x2,y2）的区域全部涂白，第二次把（x3,y3)到（x4,y4)的区域全部涂黑，问你这样以后黑白各有多少块？

题解：分割矩形，判断矩形的左下角的点是黑色还是白色就好了

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include<stack>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long LL;const int maxn = 4e4 + 10;
const int mod = 1e9 + 7;LL n,m,black,white;
int X1,X2,X3,X4,Y1,Y2,Y3,Y4;void jishu(LL lx,LL ly,LL rx,LL ry,bool flag) {LL N = ry - ly + 1, M = rx - lx + 1, b, w;LL tmp = N * M / 2;LL res = N * M - tmp;if((lx + ly) % 2) {w = tmp;b = res;}else {b = tmp;w = res;}if(flag) {white += b;black -= b;}else {black += w;white -= w;}
}void cut(LL x1,LL y1,LL x2,LL y2)
{if(x1 > x2 || y1 > y2)return;if(x2<X3 || y2<Y3 || x1>X4 || y1>Y4){jishu(x1,y1,x2,y2,1);return;}if(x1<X3) {cut(x1, y1, X3 - 1, y2);x1 = X3;}if(x2>X4) {cut(X4 + 1, y1, x2, y2);x2 = X4;}if(y1<Y3) {cut(x1, y1, x2, Y3 - 1);y1 = Y3;}if(y2>Y4) {cut(x1, Y4 + 1, x2, y2);y2 = Y4;}
}
int main()
{int t;cin >> t;while(t--){cin >> n >> m;cin >> X1 >> Y1 >> X2 >> Y2;cin >> X3 >> Y3 >> X4 >> Y4;black = n * m / 2;white = n * m - black;cut(X1,Y1,X2,Y2);jishu(X3,Y3,X4,Y4,0);printf("%lld %lld\n",white,black);}
}

 

D - Olya and magical square

 CodeForces - 1080D 

题意：有一个初始时宽为 2ⁿ的正方形，你每次可以对一个完整的正方形进行四等分。问是否存在一种方案，使得在恰好四等分 k次之后，存在一条等宽的路径，使得左下角的方块和右上角的方块联通（四联通），如果这种方案存在，输出路径的宽度对2取对数的值。

题解：n大于31的话，只需要切右下角的一块就可以了，那么答案就是n -1，n小于等于31的时候枚举答案即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<string.h>
#include<cstring>
using namespace std;
#define LL long long
const int MAXN = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int mod =  1e9 + 7;LL tot[40];
int main()
{LL n,k,t;LL cur = 1;for(int i = 0; i <= 31; i++,cur *= 4){tot[i] = (cur - 1) / 3;}cin >> t;while(t--){cin >> n >> k;if (n > 31)printf("YES %lld\n", n - 1);else{int ans = -1;for(int i = 0; i < n; i++){LL tmp = n - i,need = (1LL << tmp + 1) - tmp - 2;if(need <= k){LL last = tot[n] - ((1LL << tmp + 1) - 1) * tot[i];if(last >= k){ans = i;break;}}}if(~ans)printf("YES %d\n",ans);elseputs("NO");}}
}

 

E - Sonya and Matrix Beauty

 CodeForces - 1080E