玲珑学院-ACM比赛1014 - Absolute Defeat-白红宇的个人博客

发布日期：2021-05-09 04:21:08 浏览次数：17 分类：原创文章

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1014 - Absolute Defeat

Time Limit：2s Memory Limit：64MByte

Submissions：257Solved：73

DESCRIPTION

   Eric has an array of integers       a1,a2,...,ana1,a2,...,an. Every time, he can choose a contiguous subsequence of length       kk and increase every integer in the contiguous subsequence by       11.He wants the minimum value of the array is at least       mm. Help him find the minimum number of operations needed.  

INPUT

   There are multiple test cases. The first line of input contains an integer      TT, indicating the number of test cases. For each test case:The first line contains three integers      nn,      mm and       kk      (1≤n≤105,1≤k≤n,1≤m≤104)(1≤n≤105,1≤k≤n,1≤m≤104).The second line contains       nn integers       a1,a2,...,ana1,a2,...,an      (1≤ai≤104)(1≤ai≤104).  

OUTPUT

   For each test case, output an integer denoting the minimum number of operations needed.  

SAMPLE INPUT

    32 2 21 15 1 41 2 3 4 54 10 31 2 3 4   

SAMPLE OUTPUT

1015

源代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<stack>#include<queue>#include<vector>#include<deque>#include<map>#include<set>#include<algorithm>#include<string>#include<iomanip>#include<cstdlib>#include<cmath>#include<sstream>#include<ctime>using namespace std;typedef long long ll;int ans[200005];int main(){    int t;    int n,m,k;    int temp;    ll sum;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d",&n,&m,&k);        sum = 0;        memset(ans,0x3f3f3f3f,sizeof(ans));        for(int i = 0; i < n; i++)            scanf("%d",&ans[i]);        for(int i = 0; i < n; i++)        {            if(ans[i]<m)            {                temp=m-ans[i];                sum+=temp;                for(int j = i; j < i+k;j++)                    ans[j]+=temp;            }        }        printf("%lld\n",sum);    }	return 0;}

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