标签：map set apple problems subsets ios nes bug esc
题目链接：点击传送
You are given a set X = {1, 2, 3, 4, … , 2n-1, 2n} where n is an integer. You have to find the number of interesting subsets of this set X.
A subset of set X is interesting if there are at least two integers a & b such that b is a multiple of a, i.e. remainder of b divides by a is zero and a is the smallest number in the set.
The input file contains multiple test cases. The first line of the input is an integer T(<=30) denoting the number of test cases. Each of the next T lines contains an integer ‘n‘ where 1<=n<=1000.
For each test case, you have to output as the format below:
Case X: Y
Here X is the test case number and Y is the number of subsets. As the number Y can be very large, you need to output the number modulo 1000000007.
Input:
3
1
2
3
Output:
Case 1: 1
Case 2: 9
Case 3: 47
题意：给你2*n个数，你最小需要选两个，使得这个子集中含有最小值的倍数；
思路：枚举最小值，对于其倍数最小取一个，其余随意取与不取；
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-14 #define bug(x) cout<<"bug"<<x<<endl; const int N=1e5+10,M=1e6+10,inf=2147483647; const ll INF=1e18+10,mod=1e9+7; ll qpow(ll a,ll b,ll c) { ll ans=1; while(b) { if(b&1)ans=(ans*a)%c; b>>=1; a=(a*a)%c; } return ans; } int main() { int T,cas=1; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); ll ans=0; for(int i=1;i<=n;i++) { int p=(2*n-i); int b=((2*n)/i-1); ans=(ans+(qpow(2,p-b,mod)*(qpow(2,b,mod)+(mod-1))%mod)%mod)%mod; } printf("Case %d: %lld\n",cas++,ans); } return 0; }
标签：map set apple problems subsets ios nes bug esc
原文地址：http://www.cnblogs.com/jhz033/p/6659227.html
