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  • HDU 1003 Max Sum

    Problem Description
    Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
     
    Input
    The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
     
    Output
    For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
     
    Sample Input
    2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
     
    Sample Output
    Case 1: 14 1 4 Case 2: 7 1 6
     
    Author
    Ignatius.L
     
    题目大意:求最大字段和
    n<=100000,t<=20,可知只能使用线性复杂度的算法。
    设f[i]表示结尾为i的最大子段和,f[i]=max{f[i-1]+a[i],a[i]}
    ans=max{f[i]}
     
    #include<iostream>
    #include<cstring>
    #include<climits>
    #include<cstdio>
    using namespace std;
    const int N=100005;
    int T,n,t,a[N],f[N],res,x,y,pth[N];
    int main()
    {
        scanf("%d",&T);
        while(T--)
        {
            printf("Case %d:
    ",++t);
            scanf("%d",&n);
            for(int i=1;i<=n;i++)
                scanf("%d",a+i);
            res=f[0]=INT_MIN;
            for(int i=1;i<=n;i++)
            {
                if(f[i-1]<0)
                {
                    f[i]=a[i];
                    pth[i]=i;
                }
                else
                {
                    f[i]=a[i]+f[i-1];
                    pth[i]=pth[i-1];
                }
                if(f[i]>res)
                {
                    res=f[i];
                    x=pth[i],y=i;
                }
            }
            printf("%d %d %d
    ",res,x,y);
            if(T)
                printf("
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/fantasquex/p/10268146.html
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