HDU 1003 MAX SUM

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A - Max Sum

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1003

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

 25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

 Case 1:14 1 4Case 2:7 1 6

解题思路：HDU 1003

贪心法；

从头到尾遍历，控制好头指针和尾指针的位置即可；

AC代码如下:

#include <stdio.h>int s1[100100];int main(){    int T,n,i,s,l,r,sum,f,t;    scanf("%d",&T);    f=0;    t=T;    while(T--)    {        f++;        scanf("%d",&n);        for(i=0;i<n;i++)        {            scanf("%d",&s1[i]);        }        int sum1=s1[0];        s=0;        l=0;        r=0;        sum=0;        for(i=0;i<n;i++)        {            sum+=s1[i];            if(sum1<=sum)            {                sum1=sum;                r=s;                l=i;            }            if(sum<0)            {                s=i+1;                sum=0;            }        }        printf("Case %d:\n",f);        printf("%d %d %d\n",sum1,r+1,l+1);        if(T!=0)        {            printf("\n");        }    }    return 0;}

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