HDU 5933 ArcSoft's Office Rearrangement(构造题)

http://acm.hdu.edu.cn/showproblem.php?pid=5933

题意，给出n块价值为a的working blocks，现在要你把他们弄成k块相等的。你有两种操作，把相邻的两块合并，或者把一块拆分成任意价值的两块，问最小操作数是多少。

直接构造即可。

代码如下：

#include<bits/stdc++.h>using namespace std;typedef long long ll;ll sum, a[100005], ans;int main() {ll T, n, k, Case = 1;cin >> T;while(T--) {sum = 0;scanf("%I64d%I64d", &n, &k);for(int i = 0; i < n; i++) {scanf("%I64d", &a[i]);sum += a[i];}if(sum % k == 0) {ll ave = sum / k;ans = 0;for(int i = 0; i < n; i++) {if(a[i] < ave) {for(int j = i + 1; j < n; j++) {a[i] += a[j];ans += 1;if(a[i] == ave) {i = j;break;}else if(a[i] > ave) {ans += a[i] / ave;a[j] = a[i] - a[i] / ave * ave;if(a[j]) {i = j - 1;} else {i = j;ans--;}break;}}} else if(a[i] > ave) {ans += a[i] / ave;a[i] = a[i] - a[i] / ave * ave;if(a[i])i = i - 1;elseans--;}}printf("Case #%I64d: %I64d\n", Case++, ans);} else {printf("Case #%I64d: -1\n", Case++);}}return 0;}