(贪心)Codeforces Round #402 A. Pupils Redistribution
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题目网址: Codeforces Round #402 A. Pupils Redistribution
题意分析:
题意: 有两个小组的成绩(成绩分数范围1-5), 现要让两个组相同分数人数相等, 所以 要进行两个小组之间的人交换, 求最少的交换次数
- 思路: 分别用数组 a, 数组b记录 两个小组各个分数的人数, 然后判断某个分数的人数是否是奇数, 若是 奇数, 无法使得两个组相同分数人数相同.
- 交换次数要最少, 则交换的人数总和为(每个分数的总人数除以2 - 当前某个组该分数的人数)的绝对值.最后 交换的总人数/2(因为人是交换的)
代码:
#include <iostream>#include <cstring>#include <algorithm>using namespace std;int a[6];int b[6];int main(int argc, char const *argv[]){ int n; int tmp; int ans; while (~scanf("%d", &n)) { memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b)); for (int i = 0; i < n; ++i) { scanf("%d", &tmp); ++a[tmp]; } for (int i = 0; i < n; ++i) { scanf("%d", &tmp); ++b[tmp]; } ans = 0; for (int i = 1; i < 6; ++i) { tmp = a[i] + b[i]; if (tmp & 1) { ans = 1; break; } else { ans += abs(tmp / 2 - a[i]); } } if (ans & 1) printf("-1\n"); else printf("%d\n", ans / 2); } return 0;}
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