HDU 4704 Sum 【隔板原理+費馬小定理+快速冪】
阿新 • • 發佈:2019-01-02
Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2745 Accepted Submission(s): 1143
Problem Description
Sample Input 2
Sample Output 2 Hint 1. For N = 2, S(1) = S(2) = 1. 2. The input file consists of multiple test cases.
Source 1.隔板原理
1~N有N個元素,每個元素代表一個1.分成K個數,即在(N-1)個空擋裡放置(K-1)塊隔板
即求組合數C(N-1,0)+C(N-1,1)+...+C(N-1,N-1)。
組合數求和公式:
C(n,0)+C(n,1)+C(n,2)+.+C(n,n)=2^n
證明如下:
利用二項式定理(a+b)^n=C(n,0)a^n+C(n,1)a^(n-1)b+C(n,2)a^(n-2)b^2 +.+C(n,n)b^n
令a=b=1左邊就是2^n
所以題目即求2^(n-1)%(1e9+7)
設MOD為1e9+7
2.費馬小定理(降冪)
a^(p-1)%p=1(a,p互質)
#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<cmath> #include<queue> #include<stack> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; #define ll long long #define ms(a,b) memset(a,b,sizeof(a)) const int M=1e7+10; const int inf=0x3f3f3f3f; const int mod=1e9+7; int i,j,k,n,m; char s[M]; ll quickpow(ll a,ll b){ ll ans=1; while(b){ if(b&1) ans=(ans*a)%mod; a=a*a%mod; b>>=1; } return ans%mod; } int main() { while(~scanf("%s",s)){ int len=strlen(s); ll num=0; for(int i=0;i<len;i++){ num=(num*10+(s[i]-'0'))%(mod-1); } printf("%I64d\n",quickpow(2,num-1)); } return 0; }