acm公选课1（未完）

数的长度
Problem:65
Time Limit:1000ms
Memory Limit:65536K
Description
N! (N的阶乘) 是非常大的数，计算公式为：N! = N * (N - 1) * (N - 2) * … * 2 * 1)。现在需要知道N!有多少（十进制）位。
Input
每行输入1个正整数N。0 < N < 1000000
Output
对于每个N，输出N!的（十进制）位数。
Sample Input
1
3
32000
1000000
Sample Output
1
1
130271
5565709
Hint
Source
解题思路：
log10（100）=2
log10（199）=2.3
所以可知求阶乘的长度就是取lg的数+1
有小数的时候将double转换为int再加1

#include <bits/stdc++.h>

using namespace std;

int main()
{
    long long  n,k;
    double ans;

    while(cin>>n)
    {
        ans=0;
        for(int i=1;i<=n;i++)
        {
            ans=ans+log10(i);
        }
        k=(long long)ans+1;（变成整数型+1）
        cout<<k<<endl;
    }
    return 0;
}

最左边的数
Problem:66
Time Limit:1000ms
Memory Limit:65536K
Description
Given a positive integer N, you should output the leftmost digit of N^N.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).
Output
For each test case, you should output the leftmost digit of N^N.

Sample Input
2
3
4
Sample Output
2
2
Hint
In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2.
In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.
Source

#include <bits/stdc++.h>
typedef long long LL;
using namespace std;

int main()
{
     long long ans,m,k,n;
     double x,y;

  cin>>n;
        while(n--)
        {
            cin>>k;
            x=k*log10(k);
            y=(LL)x;
            x=x-y;
            ans=pow(10.0,x);
            cout<<ans<<endl;
        }

    return 0;
}

数字序列（循环节问题）
遥想公瑾当年，小乔初嫁了（找周期，周期不超过取余数的三倍）、、
（取余小于300考虑）
Problem:67
Time Limit:1000ms
Memory Limit:65536K
Description
A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).
Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
Output
For each test case, print the value of f(n) on a single line.

Sample Input
1 1 3
1 2 10
0 0 0
Sample Output
2
5
Hint
Source
zjcpc2004

昨日重现（1007，不能用循环节）
Problem:519
Time Limit:1000ms
Memory Limit:65536K
Description
兴安黑熊在高中学习数学时，曾经知道这样一个公式：f(n)=1²⁺²2+3^2+…+n2,这个公式是可以化简的，化简后的结果是啥它却忘记了，也许刚上大二的你能记得。现在的问题是想要计算f(n)对1007取余的值，你能帮帮他吗？
Input
输入数据有多组（不超过100组），每组一个数n. (1<=n <=1,000,000,000).
Output
输出f(n)对1007取余的值。
Sample Input
3
4
100
Sample Output
14
30
1005
Hint
本题4分，容易超时
Source
陈宇