HDU 1003 max sum

最新推荐文章于 2022-03-03 08:27:19 发布

landcruiser007

最新推荐文章于 2022-03-03 08:27:19 发布

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本文介绍了一种求解整数序列中具有最大和的子序列的算法，并提供了详细的输入输出示例及C语言实现代码。该算法适用于解决一系列与子序列最大和相关的问题。

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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

code

#include<stdio.h>
int main()
{
    int dp[100005];
    int t,k;
    while(scanf("%d",&t)!=EOF)
    {
        for(k=1;k<=t;k++)
        {
            int i,n,start=1,ft=1,lt=1;
            scanf("%d",&n);
            for(i=1;i<=n;i++)
                scanf("%d",&dp[i]);
            int max=dp[1];
            for(i=2;i<=n;i++)
            {
                if(dp[i-1]>=0)dp[i]+=dp[i-1];else start=i;
                if(dp[i]>max){max=dp[i];ft=start;lt=i;}
            }
            if(k>1)printf("\n");
            printf("Case %d:\n%d %d %d\n",k,max,ft,lt);
        }

    }
    return 0;
}