P8814 [CSP-J 2022] 解密

P8814 [CSP-J 2022] 解密

// ni = pi * qi

// ei * di = pi * qi - pi - qi + 2
// ei * di = ni - pi - qi + 2
// pi + qi = ni - ei * di + 2

// pi * qi = ni
// pi + qi = m
//
// pi * qi = 770
// pi + qi = 770 - 77*5 + 2

// m<=10^9
// m = 10000
// m/2 = 5000 5000*5000 = 25000000
// 10000 = 25000000 - ei * di + 2 = 24990002
// 24990002 = 12495001 * 2
/*
1 
25000000 12495001 2  //相同
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#define ll long long
using namespace std;
int k;
ll m, ni, di, ei, pi, qi;
int main(){
    scanf("%d",&k);
    for(int i = 1; i <= k; i++){
        scanf("%lld%lld%lld",&ni,&di,&ei);
        m = ni - ei * di + 2;
        // cout<<m/2<<endl;
        bool fl = 0 ;
        if(m % 2 == 0 && (m / 2) * (m / 2) == ni){
            fl = 1; printf("%lld %lld\n", m/2, m/2);
            continue;
        }
        if(fl == 1) continue;
        for(ll pi = 1; pi * pi <= ni; pi++){//pi小
            qi = m - pi;
            if(pi * qi == ni){
                fl = 1; printf("%lld %lld\n", pi, qi);
                break;
            }
        }
        if(!fl) printf("NO\n");
    }

    return 0;
}

温皓程二分:
x(m-x) = n
设 f(x) = -x^2 + mx
求导，得到-2x+m =0 说明是一个波峰或者波谷， 2x=m, x = m/2, 是峰或者谷
x<= m/2 有单调性，去二分它

#include<bits/stdc++.h>
using namespace std;
int T;
long long n,e,d;
int main(){
//	freopen("decode.in","r",stdin);
//	freopen("decode.out","w",stdout);
	scanf("%d",&T);
	while(T--){
		bool flag=0;
		scanf("%lld%lld%lld",&n,&e,&d);
		long long jj=e*d;
		long long cha=n-jj+2;
		long long l=1,r=cha/2;
		while(l<r){
			long long mid=(l+r+1)/2;
			if(mid*(cha-mid)>n) r=mid-1;
			else l=mid;
		}
		if(l*(cha-l)==n) printf("%lld %lld\n",min(l,(cha-l)),max(l,(cha-l)));
		else printf("NO\n");
	}
	return 0;
}