zoukankan      html  css  js  c++  java
  • 2018焦作网络赛G-Give Candies

     G: Give Candies

    时间限制: 1 Sec  内存限制: 128 MB

    题目描述

    There are N children in kindergarten. Miss Li bought them N candies。To make the process more interesting, Miss Li comes up with the rule: All the children line up according to their student number (1…N), and each time a child is invited, Miss Li randomly gives him some candies (at least one). The process goes on until there is no candy. Miss Li wants to know how many possible different distribution results are there. 

    输入

    The first line contains an integer T, the number of test case.
    The next T lines, each contains an integer N.
    1 ≤ T ≤ 100
    1 ≤ N ≤ 10^100000

    输出

    For each test case output the number of possible results (mod 1000000007).

    样例输入

    1
    4
    

    样例输出

    8

    分析

    ans=2^(n-1)

     1 #include <bits/stdc++.h>
     2 #define ll long long
     3 using namespace std;
     4 const int maxn=1e7+7;
     5 const ll mod=1e9+7;
     6 char arr[maxn];
     7 ll bit[15];
     8 ll qpow(ll a,ll b,ll c)
     9 {
    10     if(b==0)return 1;
    11     ll ans=1;
    12     while(b)
    13     {
    14         if(b&1)ans=(ans*a)%mod;
    15         a=a*a%mod;
    16         b>>=1;
    17     }
    18     return ans%mod;
    19 }
    20 ll inv(ll a,ll m)
    21 {
    22     if(a==1)return 1;
    23     return inv(m%a,m)*(m-m/a)%m;
    24 }
    25 int main()
    26 {
    27     int t;
    28     scanf("%d",&t);
    29     bit[0]=1;
    30     ll inv2=inv(2,mod);
    31     for(int i=1;i<10;i++)
    32     {
    33         bit[i]=bit[i-1]*2;
    34     }
    35     while(t--)
    36     {
    37         scanf("%s",arr);
    38         int len=strlen(arr);
    39         ll ans=1;
    40         for(int i=0;i<len;i++)
    41         {
    42             ans=qpow(ans,10,mod)%mod;
    43             ans=ans*bit[arr[i]-'0']%mod;
    44         }
    45         ans*=inv2;
    46         printf("%lld
    ",(ans+mod)%mod);
    47     }
    48     return 0;
    49 }
    View Code
     
  • 相关阅读:
    UGO小组冲刺第一天
    day04_07-三个函数的区别
    day06_08 字符串
    day06_07 字典操作02
    day06_06 字典操作01
    day06_05 字典
    day06_04 购物车讲解02
    day06_03 购物车讲解01
    day06_02 元组
    day06_01 上节回顾
  • 原文地址:https://www.cnblogs.com/CharlieWade/p/11293918.html
Copyright © 2011-2022 走看看