标签：
10
2
3
4
5
6
7
8
9
10
1000
Case 1: 1
Case 2: 1
Case 3: 2
Case 4: 2
Case 5: 2
Case 6: 2
Case 7: 4
Case 8: 12
Case 9: 12
Case 10: 744593350
思路：显然是求小于一个素数的n次方小于N的贡献为那个素数的n-1次方；
因为一次方是没用的，所以素数打表到sqrt（e14）；
求出前缀积，二分查找位置；
注意超内存跟，得到贡献那里必须要double；
#include<bits/stdc++.h> using namespace std; #define ll unsigned long long #define mod 1000000007 #define inf 100000000000005 #define MAXN 10000010 //#pragma comment(linker, "/STACK:102400000,102400000") int scan() { int res = 0 , ch ; while( !( ( ch = getchar() ) >= ‘0‘ && ch <= ‘9‘ ) ) { if( ch == EOF ) return 1 << 30 ; } res = ch - ‘0‘ ; while( ( ch = getchar() ) >= ‘0‘ && ch <= ‘9‘ ) res = res * 10 + ( ch - ‘0‘ ) ; return res ; } vector<pair<ll ,ll > >v; vector <ll>ans; ll prime[MAXN],cnt; bool vis[MAXN]; void Prime() { cnt=0; memset(vis,0,sizeof(vis)); v.push_back(make_pair(1LL,1LL)); for(ll i=2;i<MAXN;i++) { if(!vis[i]) { prime[cnt++]=i; for(double j=(double)i*i;j<inf;j*=i) v.push_back(make_pair((ll)j,i)); } for(ll j=0;j<cnt&&i*prime[j]<MAXN;j++) { vis[i*prime[j]]=1; if(i%prime[j]==0) break; } } sort(v.begin(),v.end()); ans.push_back(1LL); for(int i=1;i<v.size();i++) ans.push_back((ans[i-1]*v[i].second)%mod); } int main() { ll x,y,z,i,t; Prime(); int T,cs=1; scanf("%d",&T); while(T--) { scanf("%llu",&x); ll st=0; ll en=v.size()-1; while(st<en) { ll mid=(st+en)/2; if(v[mid].first>x) en=mid; else st=mid+1; } if(x>=v[st].first) printf("Case %d: %llu\n",cs++,ans[st]%mod); else if(x>=v[st-1].first) printf("Case %d: %llu\n",cs++,ans[st-1]%mod); } return 0; }
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原文地址：http://www.cnblogs.com/jhz033/p/5487965.html
