最大子段和-Program A

標籤：

最大子段和

Time Limit:1000MS     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<=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 4 Case 2: 7 1 6  題目大意：給出一串序列，要求輸出其子序列的最大連續和與它所在的位置。  分析：由於時間限制為一秒且n=1e5，所用兩個for迴圈肯定逾時，不過可以動態規劃，也可以二分。   代碼如下：   

#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;}

 

最大子段和-Program A