Codeforces Round #309 (Div. 2)

A. Kyoya and Photobooks
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a"to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition"photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
Input
The first line of input will be a single string s (1 ≤ |s| ≤ 20). String s consists only of lowercase English letters.
Output
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Sample test(s)
input
output
input
output

Note

In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
 

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 int main()
 5 {
 6     char str[100];
 7     int len;
 8     scanf("%s",str);
 9     len = strlen(str);
10     printf("%d
",26*(len+1)-len);
11 
12     return 0;
13 }

 
 
B. Ohana Cleans Up
Ohana Matsumae is trying to clean a room, which is divided up into an n by n grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column.
Return the maximum number of rows that she can make completely clean.
Input
The first line of input will be a single integer n (1 ≤ n ≤ 100).
The next n lines will describe the state of the room. The i-th line will contain a binary string with n characters denoting the state of the i-th row of the room. The j-th character on this line is '1' if the j-th square in the i-th row is clean, and '0' if it is dirty.
Output
The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.
Sample test(s)
input
output
input
output

Note

In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean.
In the second sample, everything is already clean, so Ohana doesn't need to do anything.
 
제목: 매번 한 열의 값(0은 1, 1은 0)을 바꿀 때마다 마지막 최대 몇 줄이 모두 1이냐고 묻는다.
사고방식: 전일이나 전영을 상관할 필요가 없다. 그들은 서로 전환할 수 있기 때문에 초기 상태, 줄마다 상태를 비교하고 가장 많은 것을 취하면 된다.
제목 링크: http://codeforces.com/contest/554/problem/B
전재는 출처를 밝혀 주십시오. 별빛 아이 찾기
 

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 
 7 char a[105][105];
 8 
 9 int main()
10 {
11     int n,i,j,ans = 0;
12     scanf("%d",&n);
13     for(i = 0;i<n;i++)
14     {
15         scanf("%s",a[i]);
16     }
17     for(i = 0;i<n;i++)
18     {
19         int  s = 0;
20         for(j = 0;j<n;j++)
21         {
22             if(strcmp(a[i],a[j])==0)
23                 s++;
24         }
25         ans = max(ans,s);
26     }
27     printf("%d
",ans);
28 }

 
 
C. Kyoya and Colored Balls
Kyoya Ootori has a bag with n colored balls that are colored with k different colors. The colors are labeled from 1 to k. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color ibefore drawing the last ball of color i + 1 for all i from 1 to k - 1. Now he wonders how many different ways this can happen.
Input
The first line of input will have one integer k (1 ≤ k ≤ 1000) the number of colors.
Then, k lines will follow. The i-th line will contain ci, the number of balls of the i-th color (1 ≤ ci ≤ 1000).
The total number of balls doesn't exceed 1000.
Output
A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1 000 000 007.
Sample test(s)
input

3
2
2
1

output

3

input

4
1
2
3
4

output

1680

Note
In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are:

1 2 1 2 3
1 1 2 2 3
2 1 1 2 3

  ：n       ， k[n] ，                    （    ）     ；       

  ：      ，    (     )     ,             <                  >，       。。。    。

    ：http://codeforces.com/contest/554/problem/C

       ：별빛 아이 찾기

            ，      。(a=1 mod (p-1),gcd(a,p)=1)

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 #define LL long long
 6 const LL mod =  1000000007;
 7 LL n;
 8 LL a[1005];
 9 LL fac[1000005];
10 
11 
12 LL ppow(LL a,LL b)
13 {
14     LL c=1;
15     while(b)
16     {
17         if(b&1) c=c*a%mod;
18         b>>=1;
19         a=a*a%mod;
20     }
21     return c;
22 }
23 
24 
25 LL work(LL m,LL i)
26 {
27     return ((fac[m]%mod)*(ppow((fac[i]*fac[m-i])%mod,mod-2)%mod))%mod;
28 }
29 
30 int main()
31 {
32     LL i,j,k;
33     fac[0] = 1;
34     for(i = 1; i<1000005; i++)
35         fac[i]=(fac[i-1]*i)%mod;
36     LL ans = 1,sum = 0;
37     scanf("%I64d",&n);
38     for(i = 1; i<=n; i++)
39     {
40         scanf("%I64d",&a[i]);
41         sum+=a[i];
42     }
43     for(i = n; i>=1; i--)
44     {
45         ans*=work(sum-1,a[i]-1);
46         ans%=mod;
47         sum-=a[i];
48     }
49     printf("%I64d
",ans);
50 
51     return 0;
52 }