【hdu 1573 X问题】【 hdu3579 Hello Kiki 】【poj 2891】

题意：求在小于等于N的正整数中有多少个X满足：X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod a[i] = b[i], … (0 < a[i] <= 10)。

//#include <bits/stdc++.h>
#include <iostream>
#define X 10005
#define inF 0x3f3f3f3f
#define PI 3.141592653589793238462643383
#define IO  ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
typedef unsigned long long Ull; //2^64
const int maxn = (int)2*1e7 + 10;
const int MOD = 9973;//(int)1e9 + 7;
const ll inf = 9223372036854775807;
//const int N = 47;
ll primer[maxn];
ll a[maxn];
int ans[maxn], num[maxn];
void ex_gcd(ll a, ll b, ll &d, ll &x, ll &y) { if (!b) { x = 1; y = 0; d = a; } else { ex_gcd(b, a%b, d, y, x); y -= x * (a / b); }; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll lcm(ll a, ll b) { return b / gcd(a, b)*a; }
ll inv_exgcd(ll a, ll m) { ll d, x, y;ex_gcd(a, m, d, x, y);return d == 1 ? (x + m) % m : -1; }
ll inv1(ll b) { return b == 1 ? 1 : (MOD - MOD / b)*inv1(MOD%b) % MOD; }    //hdu1576用这个板子会爆除0错误
ll crt(int n,int *c,int *m){ll M=1,ans=0;for(int i=0;i<n;++i) M*=m[i];
for(int i=0;i<n;++i) ans=(ans+M/m[i]*c[i] %M *inv_exgcd(M/m[i],m[i]))%M; return ans;}
ll N;
ll ex_crt(int n,ll *m,ll *c) // X ≡ c[i] mod m[i]
{
    ll M=m[0],R=c[0],x,y,d;
    for(int i=1;i<n;i++)
    {
        ex_gcd(M,m[i],d,x,y);
        if((c[i]-R)%d) return 0;
        x=(c[i]-R)/d*x%(m[i]/d);
        R+=x*M;
        M=M/d*m[i];
        R%=M;
    }
    //R: 同余方程组的最小正整数解
     R=R>0?R%M:R+M;//return (R%M+M)%M; 数据卡了0
     if(R>N)
        return 0;
     return (N-R)/M+1;//等差数列问题
}
ll f[5005];
ll sum[maxn];
ll c[maxn],m[maxn];
int cnt = 0;
int main()
{
   int t;
   ll M;
   cin>>t;
   while(t--)
   {
       cin>>N>>M;
       for(int i=0;i<M;++i)
        cin>>m[i]; //模数
        for(int i=0;i<M;++i)
            cin>>c[i]; //余数
       cout<<ex_crt(M,m,c)<<endl;
   }
    return 0;
}

HDU3579:

#include <bits/stdc++.h>
#include <iostream>
#define X 10005
#define inF 0x3f3f3f3f
#define PI 3.141592653589793238462643383
#define IO  ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
typedef unsigned long long Ull; //2^64
const int maxn = (int)2*1e7 + 10;
const int MOD = 9973;//(int)1e9 + 7;
const ll inf = 9223372036854775807;
const int N = 47;
ll c[maxn],m[maxn];
void ex_gcd(ll a, ll b, ll &d, ll &x, ll &y) { if (!b) { x = 1; y = 0; d = a; } else { ex_gcd(b, a%b, d, y, x); y -= x * (a / b); }; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll lcm(ll a, ll b) { return b / gcd(a, b)*a; }
ll inv_exgcd(ll a, ll m) { ll d, x, y;ex_gcd(a, m, d, x, y);return d == 1 ? (x + m) % m : -1; }
ll inv1(ll b) { return b == 1 ? 1 : (MOD - MOD / b)*inv1(MOD%b) % MOD; }    //hdu1576用这个板子会爆除0错误
ll crt(int n,int *c,int *m){ll M=1,ans=0;for(int i=0;i<n;++i) M*=m[i];
for(int i=0;i<n;++i) ans=(ans+M/m[i]*c[i] %M *inv_exgcd(M/m[i],m[i]))%M; return ans;}
ll ex_crt(int n,ll *m,ll *c) // R ≡ c[i] mod m[i]
{
    ll M=m[0],R=c[0],x,y,d;
    for(int i=1;i<n;i++)
    {
        ex_gcd(M,m[i],d,x,y);
        if((c[i]-R)%d) return -1;
        x=(c[i]-R)/d*x%(m[i]/d);
        R+=x*M;
        M=M/d*m[i];
        R%=M;
    }
     return R>0?R%M:R+M;//return (R%M+M)%M; 数据卡了0
}
int main()
{
    int t,n;
    cin>>t;
    for(int ca=1;ca<=t;++ca)
    {
        cin>>n;
        for(int i=0;i<n;++i)
            cin>>m[i];
        for(int i=0;i<n;++i)
            cin>>c[i];
    cout<<"Case "<<ca<<": "<<ex_crt(n,m,c)<<endl;
    }
    return 0;
}

POJ2891：

#include <iostream>
#include <cstdio>
#define X 10005
#define inF 0x3f3f3f3f
#define PI 3.141592653589793238462643383
#define IO  ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
typedef unsigned long long Ull; //2^64
const int maxn = (int)2*1e7 + 10;
const int MOD = 9973;//(int)1e9 + 7;
const ll inf = 9223372036854775807;
//const int N = 47;
ll c[maxn],m[maxn];
void ex_gcd(ll a, ll b, ll &d, ll &x, ll &y) { if (!b) { x = 1; y = 0; d = a; } else { ex_gcd(b, a%b, d, y, x); y -= x * (a / b); }; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll lcm(ll a, ll b) { return b / gcd(a, b)*a; }
ll inv_exgcd(ll a, ll m) { ll d, x, y;ex_gcd(a, m, d, x, y);return d == 1 ? (x + m) % m : -1; }
ll inv1(ll b) { return b == 1 ? 1 : (MOD - MOD / b)*inv1(MOD%b) % MOD; }    //hdu1576用这个板子会爆除0错误
ll crt(int n,int *c,int *m){ll M=1,ans=0;for(int i=0;i<n;++i) M*=m[i];
for(int i=0;i<n;++i) ans=(ans+M/m[i]*c[i] %M *inv_exgcd(M/m[i],m[i]))%M; return ans;}
ll N;
ll ex_crt(int n,ll *m,ll *c)// %m[i]=c[i]
{
    ll M=m[0],R=c[0],x,y,d;
    for(int i=1;i<n;++i)
    {
        ex_gcd(M,m[i],d,x,y);
        if((c[i]-R)%d) return -1;
        x=(c[i]-R)/d*x%(m[i]/d);
        R+=x*M;
        M=M/d*m[i];
        R%=M;
    }
     return R>0?R%M:R+M;
}
int main()
{
   int t;
   ll M;
   while(scanf("%d",&t)!=EOF){  //wa我不用多数据输入
       for(int i=0;i<t;++i)
        cin>>m[i]>>c[i];
    cout<<ex_crt(t,m,c)<<endl;
   }
    return 0;
}

【hdu 1573 X问题】【 hdu3579 Hello Kiki 】【poj 2891】

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