杭电 HDU OJ Max Sum ID1003 AC

Max Sum Accepts: 42553 Submissions: 182219
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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

#include <stdio.h>#include <stdlib.h>__int64 max_sub(int *arr,unsigned int *start,unsigned int *end,unsigned int N);int _max(int *arr,unsigned int N,unsigned int *p_start,unsigned int *p_end);int main() {    //声明变量     int T,i,*arr;    unsigned int N,j;    unsigned int start,end;    __int64 sum;    scanf("%d",&T);    //外部循环     for(i=1;i<=T;i++) {        scanf("%u",&N);        //获取N个值         arr = (int *)malloc(sizeof(int)*N);        for(j=0;j<N;j++) {            scanf("%d",&arr[j]);        }        //求解         sum = max_sub(arr,&start,&end,N);        //输出         printf("Case %d:\n",i);        printf("%I64d %u %u\n",sum,start,end);        if(T>i) printf("\n");        free(arr);    }    return 0;}__int64 max_sub(int *arr,unsigned int *p_start,unsigned int *p_end,unsigned int N) {    __int64 max=0,tmp=0;    unsigned int i,startt=0;    int flag = 1;    //循环求解     for(i=0;i<N;i++) {        //flag=1的时候重新寻找最大子串          if(arr[i]>=0&&flag==1){            startt =i+1;            flag = 0;            tmp = arr[i];        }        else if(flag==0){            tmp += arr[i];        }        //tmp小于零的时候子串丢弃         if(tmp<0){            flag=1;        }        if(max<tmp) {            max = tmp;            *p_end = i+1;            *p_start = startt;        }    }    //全为负数的情况     if(startt==0) {        return _max(arr,N,p_start,p_end);    }    else return max;}int _max(int *arr,unsigned int N,unsigned int *p_start,unsigned int *p_end) {    unsigned int i;    int max=arr[0];    *p_start = 1;    *p_end = 1;    for(i=1;i<N;i++) {        if(max<arr[i]){            max = arr[i];            *p_start=i+1;            *p_end=i+1;        }    }    return max;}