hdu5933ArcSoft's Office Rearrangement

来源：互联网发布：淘宝收货时间怎么设置编辑：程序博客网时间：2024/05/01 18:39

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

题意：将给定的n堆员工重新平均分成k堆，每次只能合并相邻两堆或者将一堆分成相邻的两堆，不可行输出-1或者输出最小操作数。

分析：很明显只需要从左往右模拟即可，比平均值多就分出一堆，小了就合并到下一堆。

代码：

#include<map>#include<set>#include<stack>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<vector>#include<string>#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef double db;typedef long long ll;typedef unsigned int uint;typedef unsigned long long ull;const db eps=1e-8;const int N=1e5+10;const int M=2e5+10;const ll MOD=1000000007;const int mod=1000000007;const int MAX=1000000010;const double pi=acos(-1.0);ll d,k,sum,a[N],ans;int main(){    int i,n,ca,T;    scanf("%d", &T);    for (ca=1;ca<=T;ca++) {        scanf("%d%I64d", &n, &k);        sum=0ll;        for (i=1;i<=n;i++) scanf("%I64d", &a[i]),sum+=a[i];        printf("Case #%d: ", ca);        if (sum%k!=0) printf("-1\n");        else {            d=sum/k;ans=0;            for (i=1;i<=n;i++)            if (a[i]<d) {                ans++;a[i+1]+=a[i];            } else if (a[i]==d) continue ;                else {                    while (a[i]>d) ans++,a[i]-=d;                    if (a[i]==d) continue ;                    else ans++,a[i+1]+=a[i];                }            printf("%I64d\n", ans);        }    }    return 0;}

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