HDU 1003（A - 最大子段和）
题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87125#problem/A
题目：
Max Sum
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
题意：
给出一个序列，求此序列的最大子段和 及开始和结束的位置。
分析：
动态规划。求最大子段和。b[i]表示最大子段和。c[i]表示子段和开始的位置。b[i]=(b[i-1]+a[i])>a[i]?b[i-1]+a[i]:a[i]
代码：
1 #include2 #include 3 using namespace std; 4 const int maxn=100005; 5 6 int a[maxn],b[maxn],c[maxn];//b表示最大子段和，c表示开始的位置 7 8 int main() 9 {10 int t,m=1;11 scanf("%d",&t);12 while(t--)13 {14 int n;15 scanf("%d",&n);16 for(int i=1;i<=n;i++)17 scanf("%d",&a[i]);18 b[1]=a[1];//记开始时的子段和为b[1]19 c[1]=1;//开始时的位置为120 for(int i=2;i<=n;i++)21 {22 if(b[i-1]>=0)23 {24 b[i]=b[i-1]+a[i];//子段和25 c[i]=c[i-1];26 }27 else28 {29 b[i]=a[i];30 c[i]=i;31 }32 }33 int max=b[1];//令起始位置最大值为b[1]34 int end=1;35 for(int i=2;i<=n;i++)36 {37 if(b[i]>max)38 {39 max=b[i];40 end=i;//结束位置41 }42 }43 printf("Case %d:\n",m++);44 printf("%d %d %d\n",max,c[end],end);45 if(t)46 printf("\n");47 }48 return 0;49 }
我先在杭电上做了几遍，提交一直都是WA，改了几次都是WA。第一次是因为我直接把开始位置写为1，根本没有计算c[i]。后来又出现了PE，因为我没有写if(t)。改了几次终于改成功了。