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
根据题目要求，每一个位置有不同的状态和，用dp[]数组记录当前的和，当从第一个数开始，依次往后累加，dp[i]记录i位置的和，当dp[i-1]<0时应该舍弃前面的和，将当前位置的arr[i]数值赋值给dp[i],然后寻找dp[i](i:1~n)中的最大值，然后记录当前的位置为end,再以当前位置往前寻找 start ,如果出现dp[i](i:end~0)<0则退出并将start赋值为i
#include<stdio.h>
int main(){
int t,n,max,start,end;
int arr[100001];
int dp[100001];
scanf("%d",&t);
for(int j=0;j<t;j++){
max =-9999;
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d",&arr[i]);
}
dp[0]=arr[0];
start=end=0;
for(int i=1;i<n;i++){
if(dp[i-1]>=0){
dp[i]=dp[i-1]+arr[i];
}else
dp[i]=arr[i];
}
for(int i=0;i<n;i++){
if(max<dp[i]){
max=dp[i];
end = i;
}
}
// start = end;// 这里是一个关键点，在这里wrong N次
for(int i=end;i>=0;i--){
if(dp[i]>=0){
start=i;
}else
break;
}
printf("Case %d:\n",j+1);
printf("%d %d %d\n",max,start+1,end+1);
if(j!=t-1)
printf("\n");
}
return 0;
}
