• 51nod“省选”模测第二场 B 异或约数和(数论分块)


    题意

    题目链接

    Sol

    这题是来搞笑的吧。。

    考虑一个数的贡献是(O(frac{N}{i}))

    直接数论分块。

    #include<bits/stdc++.h> 
    #define Pair pair<int, int>
    #define MP(x, y) make_pair(x, y)
    #define fi first
    #define se second
    #define int long long 
    #define LL long long 
    #define ull unsigned long long 
    #define Fin(x) {freopen(#x".in","r",stdin);}
    #define Fout(x) {freopen(#x".out","w",stdout);}
    using namespace std;
    const int MAXN = 1e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;
    const double eps = 1e-9;
    template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
    template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
    template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
    template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
    template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
    template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
    template <typename A> inline void debug(A a){cout << a << '
    ';}
    template <typename A> inline LL sqr(A x){return 1ll * x * x;}
    template <typename A, typename B> inline LL fp(A a, B p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;}
    template <typename A> A inv(A x) {return fp(x, mod - 2);}
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N;
    int S(int n) {
        LL t= n&3;
        if (t&1) return t/2ull^1;
        return t/2ull^n;
    }
    signed main() {
    	N = read();
    	int ans = 0;
    	for(int i = 1, nxt; i <= N; i = nxt + 1) {
    		nxt = N / (N / i); 
    		if((N / i) & 1) {
    			ans ^= S(nxt) ^ S(i - 1);
    		}
    	}
    	cout << ans;	
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10589554.html
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