codeforces比赛题A. Nineteen，BThree matrices

1、http://codeforces.com/contest/393/problem/A

题目：

A. Nineteen

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Alice likes word "nineteen" very much. She has a string
s and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.

For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw",
containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.

Help her to find the maximum number of "nineteen"s that she can get in her string.

Input
The first line contains a non-empty string s, consisting only of lowercase English letters. The length of string
s doesn't exceed 100.

Output
Print a single integer — the maximum number of "nineteen"s that she can get in her string.

Sample test(s)

Input

nniinneetteeeenn

Output

2

Input

nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii

Output

2

Input

nineteenineteen

Output

2

AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char s[105];
int c[30];
int main()
{
while(scanf("%s",s)!=EOF)
{
int len=strlen(s);
memset(c,0,sizeof(c));
for(int i=0;i<len;i++)
{
c[s[i]-'a']++;
}
int sn=c['n'-'a'];
int se=c['e'-'a'];
int si=c['i'-'a'];
int st=c['t'-'a'];
int minn=min(min(se/3,si),st);
if((sn-1)/2<minn)
printf("%d\n",(sn-1)/2);
else
printf("%d\n",minn);
}
return 0;
}

2、http://codeforces.com/problemset/problem/393/B

B. Three matrices

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Chubby Yang is studying linear equations right now. He came up with a nice problem. In the problem you are given an
n × n matrix
W, consisting of integers, and you should find two
n × n matrices A and
B, all the following conditions must hold:

Aij = Aji, for all
i, j (1 ≤ i, j ≤ n);

Bij =  - Bji, for all
i, j (1 ≤ i, j ≤ n);

Wij = Aij + Bij,
for all i, j
(1 ≤ i, j ≤ n).

Can you solve the problem?

Input
The first line contains an integer n
(1 ≤ n ≤ 170). Each of the following n lines contains
n integers. The j-th integer in the
i-th line is Wij
(0 ≤ |Wij| < 1717).

Output
The first n lines must contain matrix
A. The next n lines must contain matrix
B. Print the matrices in the format equal to format of matrix
W in input. It is guaranteed that the answer exists. If there are multiple answers, you are allowed to print any of them.

The answer will be considered correct if the absolute or relative error doesn't exceed
10 - 4.

Sample test(s)

Input

2
1 4
3 2

Output

1.00000000 3.50000000
3.50000000 2.00000000
0.00000000 0.50000000
-0.50000000 0.00000000

Input

3
1 2 3
4 5 6
7 8 9

Output

1.00000000 3.00000000 5.00000000
3.00000000 5.00000000 7.00000000
5.00000000 7.00000000 9.00000000
0.00000000 -1.00000000 -2.00000000
1.00000000 0.00000000 -1.00000000
2.00000000 1.00000000 0.00000000

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 175
double a

,b

,c

;
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
scanf("%lf",&a[i][j]);
}
for(int i=1;i<=n;i++)
{
b[i][i]=a[i][i];
for(int j=i+1;j<=n;j++)
{
b[i][j]=b[j][i]=(a[i][j]+a[j][i])/2;
c[i][j]=a[i][j]-b[i][j];
c[j][i]=-c[i][j];
}
}
for(int i=1;i<=n;i++)
{
int flag=0;
for(int j=1;j<=n;j++)
{
if(flag==0)
printf("%.8f",b[i][j]),flag=1;
else
printf(" %.8f",b[i][j]);
}
printf("\n");
}
for(int i=1;i<=n;i++)
{
int flag=0;
for(int j=1;j<=n;j++)
{
if(flag==0)
printf("%.8f",c[i][j]),flag=1;
else
printf(" %.8f",c[i][j]);
}
printf("\n");
}
}
return 0;
}