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
思路:
1.动态规划,转化方程:a[i]=max(a[i-1],a[i-1]+a[i]);
2.注意输出格式,否则Presentation Error;
#include<iostream>
using namespace std;
int a[100010],t,n,flagi,flagj,start;//flag用于记录位置
long long int sumMAX;//,dp[100010];
int main(){
int k=1,m;
cin>>t;
m=t;//m用于控制输出格式,防止演示出错
while(t--){
flagi=flagj=start=0;
cin>>n;
for(int i=0;i<n;i++)
cin>>a[i];
sumMAX=a[0];
for(int i=1;i<n;i++){
//a[i]=max(a[i-1],a[i-1]+a[i])
if(a[i-1]>=0){
a[i]=a[i]+a[i-1];
}
else{
start=i;
}
if(a[i]>sumMAX){
sumMAX=a[i];
flagi=start;
flagj= i;
}
}
cout<<"Case "<<k++<<":"<<endl;
if(m==1||m==k-1)
cout<<sumMAX<<" "<<flagi+1<<" "<<flagj+1<<endl;
else
cout<<sumMAX<<" "<<flagi+1<<" "<<flagj+1<<endl<<endl;
}
}
/*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*/