题目:
5 1 2 3 4 5
1 2 3 12 10
题意:求出LCM{C(N,0),C(N,1),....,C(N,N)}%1000000007。
思路:打表找出前几项,然后在OEIS上发现LCM{C(N,0)...C(N,N)}=LCM(1,2,3,...,N)/N,所以我们只要求出LCM{1,2,3...,N},然后再除以n即可。算LCM的时候只要找出每个小于N的质数的次数最大值,然后把他们乘起来就可以了。除以n的时候由于要取模,所以变为乘以n对与mod的逆元。
代码:
#include <cstdlib> #include <cctype> #include <cstring> #include <cstdio> #include <cmath> #include<climits> #include <algorithm> #include <vector> #include <string> #include <iostream> #include <sstream> #include <map> #include <set> #include <queue> #include <stack> #include <fstream> #include <numeric> #include <iomanip> #include <bitset> #include <list> #include <stdexcept> #include <functional> #include <utility> #include <ctime> using namespace std; #define PB push_back #define MP make_pair #define REP(i,x,n) for(int i=x;i<(n);++i) #define FOR(i,l,h) for(int i=(l);i<=(h);++i) #define FORD(i,h,l) for(int i=(h);i>=(l);--i) #define SZ(X) ((int)(X).size()) #define ALL(X) (X).begin(), (X).end() #define RI(X) scanf("%d", &(X)) #define RII(X, Y) scanf("%d%d", &(X), &(Y)) #define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z)) #define DRI(X) int (X); scanf("%d", &X) #define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y) #define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z) #define OI(X) printf("%d",X); #define RS(X) scanf("%s", (X)) #define MS0(X) memset((X), 0, sizeof((X))) #define MS1(X) memset((X), -1, sizeof((X))) #define LEN(X) strlen(X) #define F first #define S second #define Swap(a, b) (a ^= b, b ^= a, a ^= b) #define Dpoint strcut node{int x,y} #define cmpd int cmp(const int &a,const int &b){return a>b;} /*#ifdef HOME freopen("in.txt","r",stdin); #endif*/ const int MOD = 1e9+7; typedef vector<int> VI; typedef vector<string> VS; typedef vector<double> VD; typedef long long LL; typedef pair<int,int> PII; //#define HOME int Scan() { int res = 0, ch, flag = 0; if((ch = getchar()) == '-') //判断正负 flag = 1; else if(ch >= '0' && ch <= '9') //得到完整的数 res = ch - '0'; while((ch = getchar()) >= '0' && ch <= '9' ) res = res * 10 + ch - '0'; return flag ? -res : res; } /*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/ int vis[1000000+5]; int prime[1000000+5]; int cnt; void getprime() { cnt=0; MS0(vis); for(int i=2;i<=1000000;i++) { if(!vis[i]) prime[cnt++]=i; for(int j=0;j<cnt&&prime[j]<=1000000/i;j++) {vis[prime[j]*i]=1; if(i%prime[j]==0) break; } } } long long inv(long long a,long long m) { if(a==1) return 1; return inv(m%a,m)*(m-m/a)%m; } const int mod= 1e9+7; int main() {getprime(); int T; RI(T); while(T--) { int n; RI(n); n++; int ans=1; for(int i=0;i<cnt&&prime[i]<=n;i++) { //int k=(int)(log10(n)/log10(prime[i])+0.5); int tmp=1; for(;(long long )tmp*prime[i]<=n;) tmp=((long long )tmp*prime[i])%mod; ans=((long long )ans*tmp)%mod; } ans=((ans*inv(n,mod))%mod+mod)%mod; printf("%d\n",ans); } return 0; }
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原文地址:http://blog.csdn.net/u013840081/article/details/48132107