从hdu1003开始的一系列题目

hdu1003

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     分析：不是什么难题，就是让你求和最大的子数列； 代码其实不难，跑一遍应该就能懂，主要是思路就是不断变更起点。 从第一个数开始不断找出更大的子数列，并将子数列的起点和终点记录下来。 变更终点的条件是找到了一个更大的子序列，变更起点的条件是以当前起点的子序列的和已经成为负值， 最后要注意的就是，如果当前输出不是最后一个输出要输出两个换行符，即空行 #include #include #include using namespace std; int main() { int n,i,j,m; int sum,max,temp; int x,y,k; int head; cin>>n; 表示有多少组输入 for( k=1;k<=n;k++) { cin>>m>>temp; m表示一组输入有多少个，temp存储当前输入 max=sum=temp; head=0; x=y=1; for(i=1;i>temp; if(sum+tempmax) 找到了一条比当前子数列更长的子数列 { max=sum; 变更最大值 x=head+1; 变记录新数列的头尾节点 y=i+1; } } cout<<"Case"<<" "<