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
大意：
给出一个序列，求出序列中连续的数的和的最大值并输出连续数列中首尾的位置，比如6 -1 5 4 -7 中6-1+5+4=14为连续数列中和最大的。
输入：
先输入t，表示t组测试数据；
每组测试数据第一个数为序列的个数；
输出：
输出要求的最大和，和求出的序列的首尾位置。
思路：
用数组a[]记录序列中的数，对于a[i]只有两种可能 1.为一个序列的首 2.为一个序列的尾 用数组d[i]记录以第i个数结尾的序列的最大和，则
d[i]=max(d[i-1]+a[i],a[i])，d[i-1]+a[i]和a[i]分别对应a[i]的两种情况。
1 #include<cstdio> 2 #include<algorithm> 3 #include<string.h> 4 using namespace std; 5 int a[100000+11]; 6 int d[100000+11]; 7 int main() 8 { 9 int t; 10 scanf("%d",&t); 11 int k=0; 12 while(t--) 13 { 14 int sum=0; 15 int n,begin,end; 16 int max0=-1001; //max0必须小于所有可能的整数 17 scanf("%d",&n); 18 memset(a,0,sizeof(a)); 19 memset(d,0,sizeof(d)); 20 for(int i = 1 ; i <= n ; i++) 21 { 22 scanf("%d",&a[i]); 23 d[i]=max(d[i-1]+a[i],a[i]); 24 if(max0 < d[i]) 25 { 26 max0=d[i]; //记录序列和的最大值 27 end=i; //记录和最大的序列的尾 28 } 29 } 30 for(int i = end ; i >= 1 ; i--) 31 { 32 sum+=a[i]; 33 if(sum == max0) 34 { 35 begin=i; 36 } 37 38 } 39 printf("Case %d: ",++k); 40 printf("%d %d %d ",max0,begin,end); 41 if(t) 42 { 43 printf(" "); //注意输出格式 44 } 45 46 } 47 }