Codeforces Round #146 (Div. 2)

A 判断不同字母的个数为偶数还是奇数

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int main()
{
    int i,j,k,n,num[500];
    memset(num,0,sizeof(num));
    char str[201];
    gets(str);
    k = strlen(str);
    for(i = 0 ; i < k ; i++)
    num[str[i]]++;
    int count = 0;
    for(i = 0 ; i < 500 ; i++)
    if(num[i])
        count++;
    if(count%2==0)
    printf("CHAT WITH HER!\n");
    else
    printf("IGNORE HIM!");
    return 0;
}

 B题

求一个数的因数的个数公式

对一个数X分解质因子得 X = p1a1 *
p2a2 * p3a3 * ... pkak  ，X的因子数为 (a1 + 1)*(a2 + 1)*(a3 + 1)*...*(ak + 1);

写的乱七八糟的 

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
__int64 s,ss;
int num[1000001],dis[10000],pr[100000],pd[100000];
int main()
{
    int a,b,c,i,j,k;
    cin>>a>>b>>c;
    for(i = 2; i <= 10000 ; i++)
    {
        if(!pr[i])
        {
            for(j = i+i ; j <= 10000 ; j+=i)
            if(pr[j])
            pr[j] = 1;
        }
    }
    int g = 0;
    for(i =2; i <= 10000 ; i++)
    if(!pr[i])
    pd[g++] = i;
    s = 0;
    for(i = 1; i <= a ; i++)
    for(j = 1; j <= b ; j++)
    for(k = 1; k <= c ; k++)
    {
        ss = i*j*k;
        __int64 xx = ss;
        if(!num[ss])
        {
            int y = 0,kk =0 ,tt = 1;
            while(xx)
            {
                if(xx%pd[y]==0)
                {
                    kk++;
                    xx = xx/pd[y];
                }
                else
                {
                    tt = tt*(kk+1);
                    y++;
                    kk = 0;
                }
                if(xx==1)
                break;
            }
            if(kk)
            tt = tt*(kk+1);
            num[ss] = tt;
        }
        s+=num[ss];
        s = s%1073741824;
    }
    printf("%I64d\n",s);
    return 0;
}

　　C题 直接枚举 忘用——int64了

View Code 
 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define INF 0xfffffff
 6 using namespace std;
 7 __int64 gcd(__int64 a,__int64 b)
 8 {
 9     return b==0?a:gcd(b,a%b);
10 }
11 int main()
12 {
13     int i,j,k,n;
14     __int64 s = 1,max = 0;
15     cin>>n;
16     if(n>200)
17     {
18         for(i = n-200 ; i <= n ; i++)
19         for(j = i ; j <= n ; j++)
20         for(k = j ; k <= n ; k++)
21         {
22             s = i/(gcd(i,j))*j;
23             s = s/(gcd(s,k))*k;
24             if(s>max)
25             max = s;
26         }
27     }
28     else
29     {
30         for(i = 1 ; i <= n ; i++)
31         for(j = i ; j <= n ; j++)
32         for(k = j ; k <= n ; k++)
33         {
34             s = i/(gcd(i,j))*j;
35             s = s/(gcd(s,k))*k;
36             if(s>max)
37             max = s;
38         }
39     }
40     printf("%I64d\n",max);
41     return 0;
42 }

 

posted @ 2012-10-21 21:58  _雨  阅读(323)  评论(0)    收藏  举报