[PAT (Advanced Level) ]1007. Maximum Subsequence Sum 解题文档
1007. Maximum Subsequence Sum (25)
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.
Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.
Input Specification:
Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.
Output Specification:
For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.
Sample Input:10 -10 1 2 3 4 -5 -23 3 7 -21Sample Output:
10 1 4
分析：这一题有点难度。难度在于选择什么样的算法，笔者并没有找到快速的算法。笔者通过硬算的方式计算出结果。并且通过了所有测试点。
C++代码如下。
#include<iostream>
#include<vector>
using namespace std;
vector<int> v;
vector<int> sumi;
vector<int> sumv;
int main(){
int N;
cin>>N;
for(int i=0;i<N;i++){
int temp;
cin>>temp;
v.push_back(temp);
}
for(int i=0;i<N;i++){
int sum=v[i];
int big=v[i];
int big_index=i;
vector<int> sumtemp;
sumtemp.push_back(sum);
for(int j=i+1;j<N;j++){
sum+=v[j];
if(sum>big){
big=sum;
big_index=j;
}
sumtemp.push_back(sum);
}
sumv.push_back(sumtemp[big_index-i]);
sumi.push_back(big_index);
}
//find
int big=-1;
int big_index=0;
for(int i=0;i<N;i++){
if(sumv[i]>big){
big=sumv[i];
big_index=i;
}
}
if(sumv[big_index]<0){
cout<<"0 "<<v[0]<<" "<<v[N-1];
}
else
cout<<sumv[big_index]<<" "<<v[big_index]<<" "<<v[sumi[big_index]];
return 0;
}
测试点均通过。不过笔者认为这题有更有效的算法来求解，欢迎交流。