HDU 5970 CCPC2016合肥 求等差数列整除整数下取整求和

Posted 2020-10-17 HITLJR

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\n",ans);
    }
    return 0;
}