第十四届浙江财经大学程序设计竞赛重现赛B-数论

题意:让你根据公式l到r取模k的累乘，根据辗转相除法，可得如果a>k的话 gcd(a,k) = gcd(a%k,k)

那么f(a,k)%k = f(a%k,k)%k了然后我们先暴力求出1~k的f(a,k)的累乘ret，假设我们求l到r里面有m个取余k相当于1~k的区间然后这个区间的答案就是ret^m%k，接着我们还剩下l%k~k-1还有1~r%k这段区间的f(a,k)没算上去，因为k最大就10W所以我们暴力求一下加进去就好了

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int mx = 1e5+5;
typedef long long int ll;
ll l,r,k;
int a[mx];
ll gcd(ll a,ll b){
    return b==0?a:gcd(b,a%b);
}
ll modexp(ll x,ll n){
    ll ans = 1;
    while(n){
        if(n&1) ans = ans*x%k;
        x = x*x%k;
        n/=2;
    }
    return ans;
}
void solve(){
    ll L = l/k+1;
    ll R = r/k;
    ll ret = 1;
    for(int i = 1; i <= k; i++)
        ret = ret*a[i]%k;
    ll ans = modexp(ret,R-L);
    for(int i = l%k; i < k; i++)
        ans = ans*a[i]%k;
    r%=k;
    for(int i = 1; i <= r; i++)
        if(gcd(i,k)==1)
            ans = ans*(i%k)%k;
    printf("%lld\n",ans);
}
int main(){
    int t;
    scanf("%d",&t);
    int ca = 1;
    while(t--){
        scanf("%lld%lld%lld",&l,&r,&k);
        printf("Case #%d: ",ca++);
        for(int i = 1; i <= k; i++)
            if(gcd(i,k)==1)
                a[i] = i;
            else
                a[i] = 1;
        a[0] = 1;
        if(r-l+1<k){
            ll ans = 1;
            for(ll i = l; i <= r; i++)
                ans = ans*a[i%k]%k;
            printf("%lld\n",ans);
            continue;
        }
        solve();
    }
    return 0;
}