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; else ans = -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; else ans = ((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
E - Sonya and Matrix Beauty
CodeForces - 1080E
原文地址:https://www.cnblogs.com/smallhester/p/11178090.html
时间: 2024-11-14 12:51:44