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Problem Description
Find the biggest integer n (1 <= n <= N) and an integer x to make them satisfy
Input
The input consists of several test cases. Each test case contains a integer N, 1<=N<=10^18.The input ends with N = 0.
Output
In one line for each case, output two integers n and x you have found.
Sample Input
1
2
0Sample Output
1 1
1 1
题意:输入一个数 N,找到满足上式中的 n、x,且 n<=N
思路:
将右边式子的分子求和化简,有:
移项化简,有:
满足佩尔方程的形式,套入公式求解即可。
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Source Program
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-9
#define INF 0x3f3f3f3f
//#define LL long long
const int MOD=10007;
const int N=100+5;
const int dx[]= {-1,1,0,0};
const int dy[]= {0,0,-1,1};
using namespace std;
long long n[N],x[N];
vector<long long> nn,xx;
void init(){
n[0]=7LL,x[0]=1LL;
nn.push_back(1LL),xx.push_back(1LL);
for(int i=1;;i++){
n[i]=7LL*n[i-1]+48LL*x[i-1];
x[i]=n[i-1]+7LL*x[i-1];
if(n[i]<0)
break;
if((n[i]-3)%4==0){
nn.push_back((n[i]-3)/4);
xx.push_back(x[i]);
}
}
nn.push_back((long long)1e18+5);
}
int main(){
init();
long long k;
while(scanf("%lld",&k)!=EOF&&k){
for(int i=0;i<nn.size();i++){
if(k<nn[i]){
printf("%lld %lld\n",nn[i-1],xx[i-1]);
break;
}
}
}
return 0;
}