标签:
原题链接:http://codeforces.com/gym/100203/attachments/download/1702/statements.pdf
考虑到交换可以减少一次操作,那么可以的话就尽量交换,设问号的个数是z,上面是1下面是0的个数是x,反过来的是y,那么最后的答案就是z+min(x,y)+abs(x-y)。代码是队友写的,我是嘴炮流选手。
#include <stdio.h> #include <algorithm> #include <iostream> #include <string.h> #include <queue> #include <ctime> #include <cmath> #include <set> #include <stack> #include <map> #define eps 1e-10 #define MAXN 500010 #define INF 2*0x3f3f3f3f using namespace std; int T; char str1[110], str2[110]; int main() { //freopen("in.in", "r", stdin); //freopen("out.out", "w", stdout); scanf("%d", &T); for (int cas = 1; cas <= T; cas++) { scanf("%s %s", str1, str2); int len = strlen(str1); int x = 0, y = 0, z = 0; for (int i = 0; i < len; i++) if (str1[i] != ‘1‘) x++; for (int i = 0; i < len; i++) if (str2[i] != ‘1‘) y++; if (x < y) printf("Case %d: %d\n", cas, -1); else { x = y = z = 0; for (int i = 0; i < len; i++) { if (str1[i] == ‘?‘) z++; else if (str1[i] == ‘0‘ && str2[i] == ‘1‘) x++; else if (str1[i] == ‘1‘ && str2[i] == ‘0‘) y++; } printf("Case %d: %d\n", cas, z + min(x, y) + abs(y - x)); } } return 0; }
Codeforces Gym 100203E bits-Equalizer 贪心
标签:
原文地址:http://www.cnblogs.com/HarryGuo2012/p/4733117.html