題意:讓你根據公式l到r取模k的累乘,根據輾轉相除法,可得如果a>k的話 gcd(a,k) = gcd(a%k,k)
那麼f(a,k)%k = f(a%k,k)%k瞭然後我們先暴力求出1~k的f(a,k)的累乘ret,假設我們求l到r裏面有m個取餘k相當於1~k的區間然後這個區間的答案就是ret^m%k,接着我們還剩下l%k~k-1還有1~r%k這段區間的f(a,k)沒算上去,因爲k最大就10W所以我們暴力求一下加進去就好了
#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;
}