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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2327 Accepted Submission(s):
718
http://acm.hdu.edu.cn/showproblem.php?pid=4059///这题主要是公式问题，然后稍微对容斥原理计算个数，改为mult的特殊利用即可；#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<queue>using namespace std;///1~n这n个数的4次方和：n*(n+1)*(6*n*n*n+9*n*n+n-1)/30； 这里百度的公式；typedef long long ll;const ll N = 1e6+5;const ll mod = 1e9+7;ll n, ans;ll get4mult(ll mult){ return mult*mult%mod*mult%mod*mult%mod;}ll extgcd(ll a,ll b,ll &x,ll &y){ if(b==0){ x = 1, y = 0; return a; } ll d = extgcd(b,a%b,x,y); ll t = x; x = y; y = t-a/b*y; return d;}ll inverse(ll t){ ll x, y; ll d = extgcd(t,mod,x,y); return (x%mod+mod)%mod;}ll calu(ll n){ return n*(n+1)%mod*(6*n*n%mod*n%mod+9*n*n%mod+n-1)%mod*inverse(30)%mod;}ll rongchi(ll n){ ll m = n; vector<ll> v; for(ll i = 2; i*i <= m; i++){ if(m%i==0){ v.push_back(i); while(m%i==0) m/=i; } } if(m>1) v.push_back(m); ll len = v.size(); for(ll i = 1; i < (1<<len); i++){ ll mult = 1, ones = 0; for(ll j = 0; j < len; j++){ if(i&(1<<j)){ ones++; mult = mult*v[j]; } } if(ones%2) {/// 变成了 - 因为假设是原来的容斥，那么cnt指的是不互质的个数，是要被剪掉的； ///而本题：也是要减掉，但是不是计算个数，而是对每一个过程有作用，减去：减变为了加，加变为了减； ans = ans-get4mult(mult)*calu(n/mult); ans = (ans%mod+mod)%mod; }else/// 变成了 + { ans = ans+get4mult(mult)*calu(n/mult); ans = (ans%mod+mod)%mod; } } v.clear(); return ans;}int main(){ ll T; cin>>T; while(T--) { cin>>n; ans = calu(n); printf("%lld\n",rongchi(n)); } return 0;}标签：
原文地址：http://www.cnblogs.com/xiaochaoqun/p/4898689.html
