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  • 玲珑学院-ACM比赛1014

    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 (1n105,1kn,1m104)(1≤n≤105,1≤k≤n,1≤m≤104).The second line contains nn integers a1,a2,...,ana1,a2,...,an (1ai104)(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
    ",sum);
        }
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/lemonbiscuit/p/7776092.html
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