HDU-5933-ArcSoft's Office Rearrangement
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ACM模版
描述
题解
给定
很明显的贪心模拟,这场比赛好像比较钟爱贪心模拟,可是这个题好坑,因为题目中约定的数据不可能超过
代码
#include <iostream>#include <cstdio>#include <queue>using namespace std;const int MAXN = 1e5 + 10;long long N, K;long long a[MAXN];queue<long long> qi;int main(){ int T; scanf("%d", &T); for (int case_ = 1; case_ <= T; case_++) { printf("Case #%d: ", case_); scanf("%lld%lld", &N, &K); long long sum = 0; for (int i = 0; i < N; i++) { scanf("%lld", a + i); sum += a[i]; } if (sum % K) { puts("-1"); continue; } while (!qi.empty()) { qi.pop(); } long long ans = 0, unit = sum / K, t; for (int i = 0; i < N; i++) { if (!qi.empty()) { t = qi.front(); qi.pop(); a[i] += t; ans++; } if (a[i] > unit) { t = a[i] / unit; ans += t; t = a[i] % unit; if (t) { qi.push(t); } else { ans--; } } else if (a[i] < unit) { qi.push(a[i]); } } printf("%lld\n", ans); } return 0;}
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