2 条题解

  • 0
    @ 2024-3-30 20:30:20
    #include <iostream>
    #include <algorithm>
    using namespace std;
    
    typedef long long LL;
    int n;
    
    LL exgcd(LL a, LL b, LL &x, LL &y)
    {
        if(!b)
        {
            x = 1, y = 0;
            return a;
        }
        LL d = exgcd(b, a % b, y, x);
        y -= a / b * x;
        return d;
    }
    
    int main ()
    {
        cin >> n;
        bool flag = true;
        LL a1, m1;
        cin >> a1 >> m1;
        for(int i = 0; i < n - 1; i ++ )
        {
            LL a2, m2;
            cin >> a2 >> m2;
            //先用扩展欧几里得求出k1和k2
            LL k1, k2;
            LL d = exgcd(a1, a2, k1, k2);
            if((m2 - m1) % d)
            {
                flag = false;
                break;
            }
            
            k1 *= (m2 - m1) / d;
            LL t = a2 / d;
            k1 = (k1 % t + t) % t;
            
            m1 = a1 * k1 + m1;
            a1 = abs(a1 / d * a2);
        }
        if(flag)
        {
            cout << (m1 % a1 + a1) % a1;
        }
        else
            cout << "-1" << endl;
        
        return 0;
    }
    
    
    • 0
      @ 2023-9-29 14:49:25

      #include<bits/stdc++.h> #define int long long #define abs fabs #define rep(i, a, b) for(int i=a;i<b;++i) #define Rep(i, a, b) for(int i=a;i>=b;--i) using namespace std; const int N = 105; int a[N][N], eps = 1e-8; int n; inline int read(){ int s=0,w=1; char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();} while(ch>='0'&&ch<='9') s=s10+ch-'0',ch=getchar(); return sw; } void put(int x) { if(x<0) putchar('-'),x=-x; if(x>=10) put(x/10); putchar(x%10^48); } int exgcd(int a, int b, int &x, int &y){ if(!b){ x=1, y=0; return a; }else{ int d=exgcd(b, a%b, y, x); y-=a/bx; return d; } } signed main(){ ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); n=read(); int x=0, m1=read(), a1=read(); rep(i, 0, n-1){ int m2=read(), a2=read(); int k1, k2; int d=exgcd(m1, m2, k1, k2); if((a2-a1)%d){ x=-1; break; } k1=(a2-a1)/d; k1=(k1 % (m2/d) + m2/d) % (m2/d); x=k1m1+a1; int m=abs(m1/dm2); a1=k1*m1+a1; m1=m; } if(x!=-1){ x = (a1 % m1 + m1) % m1; } printf("%lld",x); return 0; }

      • 1

      信息

      ID
      115
      时间
      1000ms
      内存
      128MiB
      难度
      9
      标签
      递交数
      10
      已通过
      8
      上传者