USACO-Section1.4 Arithmetic Progressions
2017-06-10 12:19
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2017-06-10
题目大意:
一个等差数列是一个能表示成a, a+b, a+2b,…, a+nb (n=0,1,2,3,…)的数列。在这个问题中a是一个非负的整数,b是正整数。写一个程序来找出在双平方数集合(双平方数集合是所有能表示成p^2 + q^2的数的集合,其中p和q为非负整数)S中长度为n的等差数列。
样例输入:
5 7样例输出:
1 437 4
2 8
29 8
1 12
5 12
13 12
17 12
5 20
2 24
题解:
这道题打个平方数的表,最大250,算出来125000不是很大,完全可以实现。没打表直接循环肯定会超时。题中采用了pair数据结构,方便按题意排序。代码:
#include<iostream> #include<fstream> #include<cstring> #include<algorithm> using namespace std; int n , m; pair<int,int> p[10005]; int bisquare[125005]; bool checkNum[125005]; int main(){ ofstream cout("ariprog.out"); ifstream cin("ariprog.in"); cin >> n >> m; int cnt = 0; memset(bisquare , 0 ,sizeof(bisquare)); for(int i = 0;i <= m;i++){ //打表 for(int j = 0;j <= m;j++){ if(!checkNum[i * i + j * j]){ bisquare[cnt++] = i * i + j * j; checkNum[i * i + j * j] = true; } } } sort(bisquare , bisquare + cnt); int flag = 0; int i , d , k; for(i = 0;i < cnt;i++){ for(d = 1;d <= (bisquare[cnt - 1] - bisquare[0])/(n - 1);d++){ for(k = 1;k < n;k++){ if(bisquare[i] + k * d > bisquare[cnt - 1] ||!checkNum[bisquare[i] + k * d]) break; } if(k == n){ p[flag].second = bisquare[i]; p[flag].first = d; flag++; } } } sort(p , p + flag); for(int i = 0;i < flag ;i++){ cout << p[i].second <<" " << p[i].first << endl; } if(flag == 0){ cout << "NONE" << endl; } return 0; }
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