最大子段和Program A

最大子段和

Time Limit:1MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 

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<=100), then N integers followed(all the integers are between -1 and 1). 
 

 

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
 
 
题目大意:给出一串序列,要求输出其子序列的最大连续和与它所在的位置。
 
 
分析:由于时间限制为一秒且n=1e5,所用两个for循环肯定超时,不过可以动态规划,也可以二分。
 
 
 
代码如下:
 
 
 
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{

    int i,j,kase=0,n,t,b,c,d,x,sum;
    scanf("%d",&t);
    while(t--)
    {

        scanf("%d",&n);
        for(j=1;j<=n;j++)
        {
            scanf("%d",&x);
            if(j==1)
            {
                sum=b=x;
                i=c=d=1;
            }
            else
            {
                if(x>x+b)
                {
                    b=x;
                    i=j;
                }
                else
                    b+=x;
            }
            if(b>sum)
            {
                sum=b;
                c=i;
                d=j;
            }
        }
        printf("Case %d:\n",++kase);
        printf("%d %d %d\n",sum,c,d);
        if(t)
            cout<<endl;

    }
    return 0;
}
 
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