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  • HDU 5970 CCPC2016合肥 求等差数列整除整数下取整求和

    先来大宝贝

    a 公差

    b 初始项

    c 除数

    n 项数

    long long get(long long a,long long b,long long c,long long n){
        if (n<=0) return 0;
        if (n==1) return (b/c) % mod;
        long long tmp = 0;
        tmp += (a/c)%mod*((n-1)*n/2%mod)%mod;
        tmp %= mod;
        tmp += (b/c)*(n)%mod;
        tmp %= mod;
        a = a%c;
        b = b%c;
        if (a==0) return tmp;
        else return (tmp+get(c,(a*n+b)%c,a,(a*n+b)/c)) % mod;
    }

    例题

    HDU 5970

    代码

    #include <bits/stdc++.h>
    long long mod = 1e9+7;
    const double ex = 1e-10;
    #define inf 0x3f3f3f3f
    using namespace std;
    long long c;
    int f(int x,int y)
    {
         c = 0;
         int t;
         while (y>0)
         {
              c +=1;
              t = x % y;
              x = y;
              y = t;
         }
         return x;
    }
    long long get(long long a,long long b,long long c,long long n){
        if (n<=0) return 0;
        if (n==1) return (b/c) % mod;
        long long tmp = 0;
        tmp += (a/c)%mod*((n-1)*n/2%mod)%mod;
        tmp %= mod;
        tmp += (b/c)*(n)%mod;
        tmp %= mod;
        a = a%c;
        b = b%c;
        if (a==0) return tmp;
        else return (tmp+get(c,(a*n+b)%c,a,(a*n+b)/c)) % mod;
    }
    long long F(long long x, long long y){
        long long ans = 0;
        for (long long i = 1; i <= y ; i++){
            for (long long j = 0; j<i; j++){
                if (j==0&&i!=1) continue;
                long long a = j;
                if ( a == 0 ) a = 1;
                long long b = i;
                if (f(a,b) == 1){
                    long long n = (x-a)/b + 1;
                    a = a*b;
                    b = b*b;
                    ans = (ans + get(b,a,c,n)) % mod;
                }
            }
        }
        return ans;
    }
    int main()
    {
        int T;
        cin >> T;
        while (T--){
            long long x;
            long long y;
            long long p;
            scanf("%I64d%I64d%I64d",&x,&y,&p);
            mod = p;
            long long ans = 0;
            for (int i = 1 ; i <= y ; i++){
                ans = (ans + F(x/i,y/i)) % mod;
            }
            printf("%I64d
    ",ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/HITLJR/p/8039325.html
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