ACM-ICPC 2018 沈阳赛区网络预赛 G. Spare Tire (容斥原理)
可推出$a_n = n^2+n, $ 设\(S_n = \sum_{i=1}^{n} a_i\) 则 \(S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\)
需要求出\([1,N]\)中与\(M\)互质的下标的和
可以容斥计算答案,\(O(1)\)时间算出\(S_n\),需要减去与\(M\)非互质的下标\(a_i\)
对于\(M\)的每一种质因数的组合\(k\),可以求出\(b_kn = (kn)^2+kn\),则也可以\(O(1)\)得到在\(bn在[1-N]\)中的和.
根据容斥原理,奇数个质因子的组合要减去,偶数个的加回
#include <iostream>
#include <string>
#include <cmath>
#include <stdio.h>
#include <vector>
using namespace std;
const int mod= 1e9+7;
typedef long long LL;
vector<LL> fac;
LL qpow(LL a,LL n)
{
LL res=1;
while(n){
if(n&1) res = res*a %mod;
a = a*a %mod;
n>>=1;
}
return res;
}
LL solve(LL n,LL m)
{
LL rev6 = qpow((LL)6,mod-2);
LL rev2 = qpow((LL)2,mod-2);
fac.clear();
LL tmp = m;
for(LL i=2;i*i<=tmp;++i){
if(tmp%i==0){
fac.push_back(i);
while(tmp%i==0) tmp/=i;
}
}
if(tmp>1) fac.push_back(tmp);
LL res = n *(n+1) %mod *(2*n+1) %mod *rev6 %mod;
LL t = n*(n+1) %mod *rev2 %mod;
res = (res+t) %mod;
int len = fac.size();
int tot = 1<<len;
for(int i=1;i<tot;++i){
int cnt = __builtin_popcount(i);
LL tmp =1;
for(int j = 0;j<len;++j){
if(i&(1<<j)){
tmp *= fac[j];
}
}
LL nn = n/tmp;
LL pt = (nn)%mod *(nn+1)%mod *(2*nn+1) %mod*rev6%mod;
pt = pt*tmp %mod *tmp %mod;
pt = (pt+nn*(tmp+tmp*nn)%mod*rev2%mod) %mod;
if(cnt&1) res = (res-pt+mod)%mod;
else res = (res+pt)%mod;
}
return res;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
LL N,M;
while(scanf("%lld %lld",&N,&M)==2){
LL res = solve(N,M);
printf("%lld\n",res);
}
return 0;
}
为了更好的明天