Method Theorem interposer BZOJ4403 Lucas +

BZOJ4403

Flapper Act + \ (Lucas \) Theorem

Consider the total number of programs, corresponding to the \ (R-L + 1 \ ) put position \ (n-\) block things, can hold the position, as seen by the total number of programs interposer method \ (C_ {n + r-l + 1} ^ {n} -1 \)

Since \ (n, n + r- l + 1 \) range is too large, a smaller modulus, consider \ (Lucas \) Theorem

\ (Lucas \) Theorem: When \ (P \) when a prime number, \ (C_ {n-} ^ {m} \ equiv C_ {n-\ BMOD P} ^ {m \ BMOD P} * C_ {n-/ P} ^ {m / p} (\ bmod p) \)

This allows the number of combinations of the upper and lower modulus of less than, the complexity of the \ (O (T * mod) \)

Code:

#include<bits/stdc++.h>
using namespace std;
int T,n,l,r;
long long jie[1000010],fan[1000010];
const int mod=1e6+3;
long long pw(long long x,int k){
    long long ans=1;
    while(k){
        ans=(ans*x)%mod;
        x=(x*x)%mod;
        k/=2;
    }
    return ans;
}
long long rev(long long x){return pw(x,mod-2);}
long long C(int x,int y){
    return jie[x]*fan[x-y]%mod*fan[y]%mod;
}
long long Lucas(int x,int y){
    if(x<y) return 0;
    if(x<mod&&y<mod) return C(x,y);
    return Lucas(x/mod,y/mod)*Lucas(x%mod,y%mod)%mod;
}
int main(){
    scanf("%d",&T);
    jie[0]=1;
    for(int i=1;i<=1000002;i++) jie[i]=(jie[i-1]*i)%mod;
    fan[1000002]=rev(jie[1000002]);
    for(int i=1000002;i>=1;i--) fan[i-1]=(fan[i]*i)%mod;
    while(T--){
        scanf("%d %d %d",&n,&l,&r);
        printf("%lld\n",(Lucas(n+r-l+1,n)+mod-1)%mod);
    }
}