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usaco Palindromic Squares

2018年04月23日 ⁄ 综合 ⁄ 共 1416字 ⁄ 字号 评论关闭

做此题纯粹为了刷水,不过对C++了解不多,不太会,用字符串写的,效率肯定差

code

/*
ID: yueqiq
PROG: palsquare
LANG: C++
*/
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define lson l,m,rt << 1
#define rson m+1,r,rt<<1|1
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("input.txt","r",stdin)
#define writef freopen("output.txt","w",stdout)
const int maxn = 101;
const int INF = 0x3fffffff;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
const double pi = acos(-1.0);
using namespace std;
string a[20]={"0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F","G","H","I"};
string change(int n,int t){

    string ans="";
    while(n){
        int tmp=n%t;
        n/=t;
        ans.insert(0,a[tmp]);
    }
    return ans;
}
bool judge(string t){
    int len=t.length();
    FOR(i,0,len/2-1){
        if(t[i]!=t[len-1-i])
            return false;
    }
    return true;
}
int main(){
    ifstream cin( "palsquare.in" ) ;
    ofstream cout( "palsquare.out" ) ;
    int N;
    cin>>N;
    FOR(i,1,300){
        string tmp=change(i*i,N);
        if(judge(tmp)){
            string t=change(i,N);
            cout<<t<<" "<<tmp<<endl;
        }
    }
	return 0;
}

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