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  • 类欧几里德算法(洛谷 P5170

    #include <iostream>
    #include <cstdio>
    #include <queue>
    #include <algorithm>
    #include <cmath>
    #include <cstring>
    #define inf 2147483647
    #define P 998244353
    #define p(a) putchar(a)
    #define For(i,a,b) for(long long i=a;i<=b;++i)
    
    using namespace std;
    long long T;
    long long n,a,b,c;
    long long i2 = 499122177, i6 = 166374059;//这是2,6在%P意义下的逆元
    struct data{
        long long f,g,h;
        data calc(long long n,long long a,long long b,long long c){
             long long ac = a / c, bc = b / c, m = (a * n + b) / c, n1 = n + 1, n21 = n * 2 + 1;
             data d;
             if (a == 0) {
                d.f = bc * n1 % P;
                d.g = bc * n % P * n1 % P * i2 % P;
                d.h = bc * bc % P * n1 % P;
                    return d;
            }
    
            if (a >= c || b >= c){
                d.f = n * n1 % P * i2 % P * ac % P + bc * n1 % P;
                d.g = ac * n % P * n1 % P * n21 % P * i6 % P + bc * n % P * n1 % P * i2 % P;
                d.h = ac * ac % P * n % P * n1 % P * n21 % P * i6 % P +
                bc * bc % P * n1 % P + ac * bc % P * n % P * n1 % P;
                d.f %= P, d.g %= P, d.h %= P;
    
                data e = calc(n, a % c, b % c, c);
    
                d.h += e.h + 2 * bc % P * e.f % P + 2 * ac % P * e.g % P;
                d.g += e.g, d.f += e.f;
                d.f %= P, d.g %= P, d.h %= P;
                    return d;
            }
    
            data e = calc(m - 1, c, c - b - 1, a);
            d.f = n * m % P - e.f, d.f = (d.f % P + P) % P;
            d.g = m * n % P * n1 % P - e.h - e.f, d.g = (d.g * i2 % P + P) % P;
            d.h = n * m % P * (m + 1) % P - 2 * e.g - 2 * e.f - d.f;
            d.h = (d.h % P + P) % P;
            return d;
        }
    }ans;
    
    void in(long long &x){
        long long y=1;char c=getchar();x=0;
        while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
        while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
        x*=y;
    }
    void o(long long x){
        if(x<0){p('-');x=-x;}
        if(x>9)o(x/10);
        p(x%10+'0');
    }
    
    signed main(){
        in(T);
        while(T--){
            in(n);in(a);in(b);in(c);
            ans=ans.calc(n,a,b,c);
            o(ans.f);p(' ');o(ans.h);p(' ');o(ans.g);p('
    ');
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/war1111/p/11303654.html
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