UVA - 12716 GCD XOR（GCD等于XOR）（数论）

时间：2023-03-10 04:18:07
UVA - 12716 GCD XOR（GCD等于XOR）（数论）

题意：输入整数n(1<=n<=30000000)，有多少对整数(a, b)满足：1<=b<=a<=n，且gcd(a,b)=a XOR b。

分析：因为c是a的约数，所以枚举c，a = k*c，通过a-c求b，并通过a^b=c来验证。

#pragma comment(linker, "/STACK:102400000, 102400000")

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<cmath>

#include<iostream>

#include<sstream>

#include<iterator>

#include<algorithm>

#include<string>

#include<vector>

#include<set>

#include<map>

#include<stack>

#include<deque>

#include<queue>

#include<list>

#define Min(a, b) ((a < b) ? a : b)

#define Max(a, b) ((a < b) ? b : a)

const double eps = 1e-8;

inline int dcmp(double a, double b){

    if(fabs(a - b) < eps) return 0;

    return a > b ? 1 : -1;

}

typedef long long LL;

typedef unsigned long long ULL;

const int INT_INF = 0x3f3f3f3f;

const int INT_M_INF = 0x7f7f7f7f;

const LL LL_INF = 0x3f3f3f3f3f3f3f3f;

const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;

const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};

const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};

const int MOD = 1e9 + 7;

const double pi = acos(-1.0);

const int MAXN = 30000000 + 10;

const int MAXT = 10000 + 10;

using namespace std;

int ans[MAXN];

void init(){

    for(int c = 1; c <= (MAXN >> 1); ++c){

        for(int a = c + c; a <= MAXN; a += c){

            int b = a - c;

            if((a ^ b) == c) ++ans[a];

        }

    }

    for(int i = 2; i <= MAXN; ++i){

        ans[i] += ans[i - 1];

    }

}

int main(){

    init();

    int T;

    scanf("%d", &T);

    int kase = 0;

    while(T--){

        int n;

        scanf("%d", &n);

        printf("Case %d: %d\n", ++kase, ans[n]);

    }

    return 0;

}

　　

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