hdu 1003 Max Sum 去掉数组后的方法

Max Sum

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 104   Accepted Submission(s) : 31

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

 

#include<stdio.h>int a[100010];int main(){    int n,i,start,end,sum,max,sta,ti,k; scanf("%d",&ti);  for(k=1;k<=ti;k++)  {   sum=0;max=-1111;   scanf("%d",&n);   for(i=1;i<=n;i++)    scanf("%d",&a[i]);   start=sta=end=1;   for(i=1;i<=n;i++)   {    sum=sum+a[i];    if(sum>max)//这个if要放在下面的if语句的上面     {     max=sum;     end=i;start=sta;    }    if(sum<0)//这个要放在下面  因为要是一开始输入的只有一个负数的时候  就会搞错了 WA了好几次 都是因为这里    {     sta=i+1; sum=0;    }   }   printf("Case %d:\n",k);   printf("%d %d %d\n",max,start,end);   if(k!=ti) printf("\n");  }  return 0;}

 

#include <stdio.h>int main(){    int i,j,m,n,max;    int s,a,b;    int x1,x2;    while(scanf("%d",&n)!=EOF&&n!=0)    {                scanf("%d",&s);        max=s;        x1=s;        x2=s;        int x3,x4;        x3=x4=s;        a=s;        for(i=2;i<=n;i++)        {            scanf("%d",&m);            b=m;            if(s+m<m)            {                x1=x2=m;                s=m;            }            else            {                x2=m;                s=s+m;            }            if(max<s)            {                x3=x1;                x4=x2;                max=s;            }        }        if(max<0)        {            max=0;            x3=a;            x4=b;        }        printf("%d %d %d\n",max,x3,x4);    }        return 0;}