zoj 2520 Amicable Pairs

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1520

 

好题,又学习了一种方法.

此题只要建一个表,里面放上它的所有因数就可以了.

新建一个因数表并打印1到500的每个数的因数:

/* * 分解因数.cpp * * Created on: Apr 6, 2010 * Author: wyy *分解因数. */#define SIZE 501#include<iostream>#include<vector>using namespace std;int main(){//freopen("input.txt", "r", stdin);vector< vector<int> > vec(SIZE);for(int k = 1; k != SIZE; ++k)vec[k].push_back(1);for(int i = 2; i != SIZE / 2; ++i){int j = 2 * i;while(j < SIZE){vec[j].push_back(i);j += i;}}for(int k = 2; k != SIZE; ++k)vec[k].push_back(k);for(int k = 1; k != SIZE; ++k){cout << k << " : ";for(vector<int>::iterator iter = vec[k].begin(); iter != vec[k].end(); ++iter)cout << *iter << " ";cout << endl;}//fclose(stdin);return 0;}

 

将因数表中的每个数的每个因数相加(题目中说除了1和本身.所以上面的程序再删除几句代码就可以了),并放到一个数组sum里面

然后判断sum[sum[i]] == i 是否成立就可以了.

好了,贴上代码:

/* * 2520.cpp * * Created on: Apr 5, 2010 * Author: wyy * */#include<iostream>#include<vector>using namespace std;struct Pairs{int x;int y;};const int SIZE = 8000000;const int NUM = 6000000;int sum[SIZE];int main(){for (int i = 2; i != NUM; i++)sum[i] = 1;for(int i = 2; i != NUM / 2; ++i){int x = 2 * i;while(x < NUM){sum[x] += i;x += i;}}Pairs P;vector<Pairs> vec;for(int i = 220; i != NUM; ++i){if(sum[i] > i && sum[i] < SIZE && sum[sum[i]] == i){P.x = i;P.y = sum[i];vec.push_back(P);}}//freopen("input.txt", "r", stdin);int k;while(cin >> k)cout << vec[k - 1].x << " " << vec[k - 1].y << endl;return 0;}