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
有2种方法做，我开始不明白题目意思，走了很多弯路。
题目要求是这样的：
输出数组的子序列的最大值。
如果有相同的，开始序列输出小的。
结尾序列输出大的。
/**1003题**/ #include <stdio.h> #include <stdlib.h> int a[1000020]; long long int c[3000020],b[1000020]; int main() { int t,tt=1; scanf("%d",&t); while(t--) { int i,j,n,a1,a2; scanf("%d",&n); int age=0; if(n!=1) { for(i=0; i<n; i++) scanf("%d",&a[i]); long long int sum=0; for(i=0; i<n; i++) { if(a[i]>=0) { a1=a2=i+1; age=1; break; } } if(age!=1) { int max=a[0]; a1=a2=1; for(i=1; i<n; i++) { if(a[i]>max) { max=a[i]; a1=a2=i+1; } } printf("Case %d:\n",tt); tt++; printf("%d %d %d\n",max,a1,a2); if(t!=0) printf("\n"); } else { int k=0; c[k]=0; long long int max=c[k]; for(i=a1-1;i<n;i++) { c[k]=0; for(j=i;j<n;j++) { c[k]=a[j]+c[k]; if(c[k]<0) break; if(max<c[k]) { a1=i+1; a2=j+1; max=c[k]; } if(max==c[k]) { if(a1>i+1) a1=i+1; if(a2<j+1) a2=j+1; } } k++; } printf("Case %d:\n",tt); tt++; printf("%I64d %d %d\n",max,a1,a2); if(t!=0) printf("\n"); } } if(n==1) { scanf("%d",&a[0]); { printf("Case %d:\n",tt); tt++; printf("%d 1 1\n",a[0]); if(t!=0) printf("\n"); } } } return 0; }
