Max Sum

2019-11-10 17:27:23

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Given a sequence a11,a22,a33......ann, 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.InputThe 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).OutputFor 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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5Sample OutputCase 1:14 1 4Case 2:7 1 6#include<stdio.h>#include<string.h>#include<math.h>int a[100010];int main(){    int t;    scanf("%d",&t);    int g=0;    while(t--)    {        ++g;        int n;        scanf("%d",&n);        memset(a,0,sizeof(a));        for(int i=1;i<=n;i++)        {         scanf("%d",&a[i]);        }        int sum=0;        PRintf("Case %d:/n",g);        int p=1,l=1,k=1;        int maxn=-1e9;        for(int i=1;i<=n;i++)        {                sum+=a[i];                if(maxn<sum)                {                    maxn=sum;                    p=i;//记序列的右边位置                    k=l;//记序列的左边位置                }           if(sum<0)            {                l=i+1;                sum=0;            }        }    printf("%d %d %d/n",maxn,k,p);    if(t)        printf("/n");        //}    }}

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