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最大连续子序列&&MAX SUM
//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 }
最大连续子序列&&MAX SUM