Description:
Have you ever Googled the “Animate”?Tyr Googles the “Animate” one day, and find out 245 match records.
But he don’t know whether the number 245 is correct or not. Here comes the problem A: Give you a
documentation,your task is to find out whether this documentation contains the word “Animate” or not.

Input:
The input will consist of several case of input.The first line is an integer N which give the case number.
The next part has N cases.Every case follows this format: Line 1: The beginning of the documentation: #Doc
Line 2 to m-1: The contest of documentation,this may be consists of several lines.Every line contains less
than 100 characters,and the total characters of one documentation is less than 1000. Line m:The end of the
documentation:#End
Output:
For each input case,your program should print an anwser in a single line.If the documentation contains
“Animate”, please print “Yes”, else print “No”. In Google, we know the match is case insensitive,
which means “animate” is also a match of “Animate”. In this problem we just take care of the alphabet
characters, please ignore other characters.

*/
    最开始，这道题我一看，非常容易可以用字符串string类的查找方法，如是写了如下的一个程序：

#include<iostream>
#include<string>
using namespace std;

int main()
{
    char tmp[101];
    string str;
    int N;
    scanf("%d%*c",&N);
    while(N--){
        bool flag = false;
        gets(tmp);
        str = tmp;
        while(str!="#END"){
            gets(tmp);
            strupr(tmp);
            str = tmp;
            //cout<<str<<endl;
            if(str.find("ANIMATE")!=4294967295){
                flag = true;
            }
        }
        if(flag){
            cout<<"Yes"<<endl;
        }else{
            cout<<"No"<<endl;
        }
    }
    return 0;
}

    结果提交的结果，为wrong answear，百思不得其解，如是在网上搜了一下，发现题目的最后一句话：只考虑字母的情况！！这句话才是解这道题目的关键。正确的代码如下所示：

#include<iostream>
#include<fstream>
#include<sstream>
using namespace std;
//-----------------------------------
int main(){
    int a;cin>>a;
    for(int i=0;i<a;i++){
        bool isyes=0;
        string s;bool head=0;
        string s2="";
        while(s!="#End"){
            getline(cin,s);
            if(s=="#Doc")
                head=1;
            if(head){
                for(int i=0;i<s.length();i++)
                    if(isalpha(s[i]))
                        s2+=tolower(s[i]);
                if(s2.find("animate")!=-1 && isyes==0){
                    cout<<"Yes"<<endl;
                    isyes=1;
                }
            }
        }
        if(isyes==0)
            cout<<"No"<<endl;
    }
}

Description:
给定一个数，其值用A的B次方表示(B<100000)，求该数除以一个整数C（<100000）所得的余数。注意算法的合理性，其性能有一定的要求。每行有三个数，依次表示A，B，C，每行对输出对应的余数。

Sample Input:
1 2 3
2 1 3
3 3 5
Sample Output:
1
2
2

#include<iostream>
using namespace std;

long power(long A,long B,long C)
{
    if(B==0){
        return 1;
    }
    if(B==1){
        return A%C;
    }

    long tmp = power(A,B/2,C);
    if(B&1==1){
        return (((tmp*tmp)%C)*(A%C))%C;
    }else{
        return (tmp*tmp)%C;
    }
    return 0;
}

int main()
{
    long A,B,C;
    while(scanf("%ld%ld%ld",&A,&B,&C)!=EOF){
        printf("%ld\n",power(A,B,C));
    }
    return 0;
}


/**//*long long PowerMod(long long a, int b, int k)  
{   //(a^b )%k   
    long long tmp = a, ret = 1;  
    for(; b ; b>>1 )  
    {  
        if (b & 1)  
            ret = (ret * tmp) % k;  //只是每步运算的时候都取模   
        tmp = (tmp * tmp) % k;  
    }  
    return ret;  
}*/

简单的事情
Time Limit:1000MS  Memory Limit:32768K

Description:
数学天才fans曾经说过一句话：组合数的计算是一件非常简单的事情。组合数的计算真的是一件非常简单的事情吗？请你自己去尝试一下吧！

Input:
输入中的一些整数对n，m（m≤n≤20）
Output:
输出其组合数。
Sample Input:
5 2
18 13
Sample Output:
10
8568

#include<iostream>
using namespace std;

int c[21][21];

int main()
{
    int n,m;
    for(int i=0;i<=20;i++){
        c[i][0] = 1;
    }
    for(int i=1;i<=20;i++){
        c[i][i] = 1;
    }
    for(int i=2;i<=20;i++){
        for(int j=1;j<i;j++){
            c[i][j] = c[i-1][j-1]+c[i-1][j];
        }
    }

    while(scanf("%d%d",&n,&m)!=EOF){
        printf("%d\n",c[n][m]);
    }
    return 0;
}