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//错的莫名其妙的O w O
第二个的格式也是莫名其妙的
Input
测试输入包含若干测试用例，每个测试用例占2行，第1行给出正整数K( < 10000 )，第2行给出K个整数，中间用空格分隔。当K为0时，输入结束，该用例不被处理。
Output
对每个测试用例，在1行里输出最大和、最大连续子序列的第一个和最后一个元
素，中间用空格分隔。如果最大连续子序列不唯一，则输出序号i和j最小的那个（如输入样例的第2、3组）。若所有K个元素都是负数，则定义其最大和为0，输出整个序列的首尾元素。
Sample Input
6
-2 11 -4 13 -5 -2
10
-10 1 2 3 4 -5 -23 3 7 -21
6
5 -8 3 2 5 0
1
10
3
-1 -5 -2
3
-1 0 -2
0
Sample Output
20 11 13
10 1 4
10 3 5
10 10 10
0 -1 -2
0 0 0
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
1 #include <stdio.h> 2 #include <string.h> 3 int main() 4 { 5 int first,last,temp,n,i,j,flag,thissum; 6 int a[22222]; 7 while(scanf("%d",&n)&&n) 8 { 9 flag=0; 10 thissum=0; 11 memset(a,0,sizeof(a)); 12 for(i=1;i<=n;i++) 13 { 14 scanf("%d",&a[i]); 15 if(a[i]>=0) 16 flag=1; 17 } 18 first=last=temp=1; 19 if(!flag) 20 { 21 printf("0 %d %d\n",a[1],a[n]); 22 continue; 23 } 24 int max=-33333; 25 for(i=1;i<=n;i++) 26 { 27 thissum+=a[i]; 28 if(thissum>max) 29 { 30 max=thissum; 31 first=temp; 32 last=i; 33 } 34 if(thissum<0) 35 { 36 thissum=0; 37 temp=i+1; 38 } 39 } 40 printf("%d %d %d\n",max,a[first],a[last]); 41 } 42 return 0; 43 }
1 #include<stdio.h> 2 int main() 3 { 4 int n,m,i,a,t=1; 5 scanf("%d",&n); 6 for(int t=1;t<=n;t++) 7 { 8 int maxsum=-0xfffffff; 9 int thissum=0; 10 int first=1,last=1; int temp=1; 11 scanf("%d",&m); 12 for(i=1;i<=m;i++) 13 { 14 scanf("%d",&a); 15 thissum+=a; 16 if(maxsum<thissum) 17 { 18 maxsum=thissum; 19 first=temp; 20 last=i; 21 } 22 if(thissum<0) 23 { 24 thissum=0; 25 temp=i+1; 26 } 27 } 28 printf("Case %d:\n",t); 29 printf("%d %d %d\n",maxsum,first,last); 30 if(t!=n) 31 printf("\n"); 32 } 33 return 0; 34 }
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原文地址：http://www.cnblogs.com/awsent/p/4268555.html
