A - Petya and Origami

?CodeForces - 1080A?

題意：制造一份邀請函需要2份a物品，5份b物品，8份c物品，一個盒子里面有k份物品（可以為a或b或c）問你制造n份邀請函需要用多少個盒子

題解：加起來就行了

#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; } View Code

?

B - Margarite and the best present

?CodeForces - 1080B? 題意：區(qū)間內(nèi)偶數(shù)和減去奇數(shù)和 題解：分類一下就好了 #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;}} View Code

?

C - Masha and two friends

?CodeForces - 1080C?

題意：給你一個n行m列的黑白塊相間的棋盤，進(jìn)行兩次操作，第一次把（x1,y1）到（x2,y2）的區(qū)域全部涂白，第二次把（x3,y3)到（x4,y4)的區(qū)域全部涂黑，問你這樣以后黑白各有多少塊？

題解：分割矩形，判斷矩形的左下角的點是黑色還是白色就好了

#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);} } View Code

?

D - Olya and magical square

?CodeForces - 1080D? 題意：有一個初始時寬為 2ⁿ的正方形，你每次可以對一個完整的正方形進(jìn)行四等分。問是否存在一種方案，使得在恰好四等分 k次之后，存在一條等寬的路徑，使得左下角的方塊和右上角的方塊聯(lián)通（四聯(lián)通），如果這種方案存在，輸出路徑的寬度對2取對數(shù)的值。 題解：n大于31的話，只需要切右下角的一塊就可以了，那么答案就是n -1，n小于等于31的時候枚舉答案即可 #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");}} } View Code

?

E - Sonya and Matrix Beauty

?CodeForces - 1080E?

轉(zhuǎn)載于:https://www.cnblogs.com/smallhester/p/11178090.html

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