元素 integer [1] bsp more nta rip tle int

Max Sum

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 代碼如下： 原理：找出最大串的方法：和並法。遍歷的同時將前面的加起來，具體看代碼(AC)

#include<stdio.h>
#include
 
<string.h>
int main()
{
    int t,num=0;;
    scanf("%d",&t);
    while(t--)
    {
        num++;
        int n,temp=1,a;//temp表示當前最大區間開頭所在位置，一開始假定為1
        scanf("%d",&n);
        int start=1,max=-1001,end1,sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d"
 
,&a);//a就是區間元素了，由於只用一次，所以就不用數組浪費內存了
            sum+=a;
            if(sum>max)//max 表示目前為止最大區間的和，sum大於max，說明區間改變
            {
              max=sum;
              end1=i+1;
              start=temp;
            }
            if(sum<0)//sum小於0，則表示若前面的temp為開頭，不會有最大區間，故將sum，temp重置。
            {
                sum
 
=0;//sum=0表示重新開始計算最大區間（下面的也是這個意思）
                temp=i+2;
            }    
        }    
        printf("Case %d:\n%d %d %d\n",num,max,start,end1);
        if(t!=0)
          printf("\n");                       
    }
    return 0;
} 

POJ 1003 Max Sum