• BZOJ3834 [Poi2014]Solar Panels 【数论】


    题目链接

    BZOJ3834

    题解

    容易想到对于(gcd(x,y) = D)(d)的倍数一定存在于两个区间中
    换言之

    [lfloor frac{a - 1}{D} floor < lfloor frac{b}{D} floor ]

    [lfloor frac{c - 1}{D} floor < lfloor frac{d}{D} floor ]

    整除分块即可做到(O(nsqrt{max{b}}))

    #include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<cmath>
    #include<map>
    #define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
    #define REP(i,n) for (int i = 1; i <= (n); i++)
    #define mp(a,b) make_pair<int,int>(a,b)
    #define cls(s) memset(s,0,sizeof(s))
    #define cp pair<int,int>
    #define LL long long int
    using namespace std;
    const int maxn = 100005,maxm = 100005,INF = 1000000000;
    inline int read(){
    	int out = 0,flag = 1; char c = getchar();
    	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    	return out * flag;
    }
    int main(){
    	register int T = read(),a,b,c,d,M,ans,A,B,C,D;
    	while (T--){
    		a = read() - 1; b = read(); c = read() - 1; d = read();
    		M = max(b,d); ans = 1;
    		for (register int i = 1,nxt; i <= M; i = nxt + 1){
    			nxt = INF;
    			A = a / i,B = b /i,C = c / i,D = d / i;
    			if (A) nxt = min(nxt,a / A);
    			if (B) nxt = min(nxt,b / B);
    			if (C) nxt = min(nxt,c / C);
    			if (D) nxt = min(nxt,d / D);
    			if (A < B && C < D) ans = nxt;
    		}
    		printf("%d
    ",ans);
    	}
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Mychael/p/9246698.html
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