UVa 11609

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先找规律，然后猜结果，AC。。

后来看了题解，证明忘了，自己给了个证明，不是很巧妙

所求为和式 k*c(n,k) k为[1,n]

它等于了和式 c(n,k)*(n-k) k为[0,n-1]

两式相加变成了和式 c(n,k)*k k为[1,n-1] +c(n,n)*n+c(n,0)*n

即 n*∑ c(n,k) k为[0,n]

组合的意义是从n个数中选k个

所以上式为n*2^n

然后除以2

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <ctime>#include <cmath>#include <vector>#include <deque>#include <algorithm>#include <list>#include <map>#include <set>#include <queue>#include <stack>#define REP(i,n) for(int i=0;i<(n);i++)#define REP1(i,n) for(int i=1;i<=(n);i++)#define FOR(i,a,b) for (int i=(a);i<=(b);i++)#define CLR(x,n) memset(x,n,sizeof(x))#define PN printf("\n")#define read(x) scanf("%d",&x)#define read2(x,y) scanf("%d%d",&x,&y)#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define write(x) printf("%d",x)#define write1(x) printf("%d ",x)#define writeln(x) printf("%d\n",x)#define write2(x,y) printf("%d %d",x,y)#define writeln2(x,y) printf("%d %d\n",x,y)#define write3(x,y,z) printf("%d %d %d",x,y,z)#define writeln3(x,y,z) printf("%d %d %d\n",x,y,z)#define LL long long#define MOD 1000000007using namespace std;/*WA 输入4会输出16。。。快速幂写错 if p%2==1... */LL quickpow(LL p,LL q){    if (q==1)return p;    if (q==0)return 1;    LL ans=quickpow(p,q/2);    ans=(ans*ans)%MOD;    if (q%2==1)ans=(ans*p)%MOD;    return ans;}int main(){    int t;read(t);LL n;    REP(k,t){        scanf("%lld",&n);        LL ans=( (n % MOD)*quickpow(2,n-1) ) % MOD;        printf("Case #%d: %lld\n",k+1,ans);    }    system("pause");    return 0;}