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
Author Ignatius.L
Recommend We have carefully selected several similar problems for you: 1176 1069 2084 1058 1203 AC代码： #include<cstdio> int a[100000+10]; int main() { int t,n,k=1; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); int max=-1001,st,endd,sum=0,st1=1; //注意整数的范围 for(int i=1;i<=n;i++) { sum+=a[i]; if(sum>max) { max=sum; st=st1; endd=i; } if(sum<0) { sum=0; st1=i+1; //st1是临时的开始点，如果后面的sum<max，这个开始点也就不会计入 } } printf("Case %d:\n%d %d %d\n",k++,max,st,endd); if(t!=0) { printf("\n"); } } return 0; }
