01月05, 2019

hdu1003

题目描述:

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.

输入:

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).

输出:

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.

样本输入:

2

5 6 -1 5 4 -7

7 0 6 -1 1 -6 7 -5

样本输出:

Case 1:

14 1 4

Case 2:

7 1 6

要求最大的sum[i],可以求sum[i-1],比较sum[i-1]+a[i]与a[i]的大小,典型的动态规划问题。

代码:

#include <stdio.h>

int main()
{
    int n,m,t;
    int sum,max;
    int x,y,X,Y;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&m);
        sum=-1001;
        max=-1001;
        for(int j=1;j<=m;j++){
            scanf("%d",&t);
            if(sum+t<t){
                sum=t;
                x=j;
                y=j;
            }
            else{
                sum+=t;
                ++y;
            }
            if(max<sum){
                max=sum;
                X=x;
                Y=y;
            }
        }
        printf("Case %d:\n",i);
        printf("%d %d %d\n",max,X,Y);
        if(i!=n)
            printf("\n");
    }
    return 0;
}

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