NCEPU第六次积分赛

D - D

Submit Status Practice ZOJ 1151

Description

For eachlist of words, output a line with each word reversed without changing the orderof the words.

This problem contains multiple test cases!

Thefirst line of a multiple input is an integer N, then a blank line followed by Ninput blocks. Each input block is in the format indicated in the problemdescription. There is a blank line between input blocks.

Theoutput format consists of N output blocks. There is a blank line between outputblocks.

 

Input

You willbe given a number of test cases. The first line contains a positive integerindicating the number of cases to follow. Each case is given on a linecontaining a list of words separated by one space, and each word contains onlyuppercase and lowercase letters.

 

Output

For eachtest case, print the output on one line.

 

Sample Input

1

3
I am happy today
To be or not to be
I want to win the practice contest

 

Sample Output

I mayppah yadot
oT eb ro ton ot eb
I tnaw ot niw eht ecitcarp tsetnoc

#include"stdio.h"int main(){   char s[1005];   int N,i,j,T,k;   scanf("%d",&T);   while(T--)   {   scanf("%d",&N);   getchar();   while(N--)   {gets(s);i=j=0;while(s[j]){if(s[j]==' '){for(k=j-1;k>=i;k--)printf("%c",s[k]);printf(" ");i=j+1;}j++;}for(k=j-1;k>=i;k--)printf("%c",s[k]);printf("\n");   }   if(T)   printf("\n");   }return 0;}

E - E

Time Limit:2000MS     MemoryLimit:65536KB     64bit IO Format:%lld& %llu

Submit Status Practice ZOJ 2736

Description

Thedaffodil number is one of the famous interesting numbers in the mathematicalworld. A daffodil number is a three-digit number whose value is equal to thesum of cubes of each digit.

Forexample. 153 is a daffodil as 153 = 1³ + 5³ + 3³.

Input

Thereare several test cases in the input, each case contains a three-digit number.

Output

One linefor each case. if the given number is a daffodil number, then output"Yes", otherwise "No".

Sample Input

153
610

Sample Output

Yes
No

#include"stdio.h"int main(){   int N,tn,sum,t;   while(scanf("%d",&N)!=EOF)   {   t=N%10;   tn=N/10;   sum=t*t*t;   t=tn%10;   tn=tn/10;   sum+=t*t*t;   t=tn%10;   tn=tn/10;   sum+=t*t*t;   printf("%s\n",sum==N?  "Yes":"No");         }return 0;}

F - F

Time Limit:2000MS     MemoryLimit:65536KB     64bit IO Format:%lld& %llu

Submit Status Practice ZOJ 2850

Description

Tom's Meadow

Tom hasa meadow in his garden. He divides it into N * M squares.Initially all the squares were covered with grass. He mowed down the grass onsome of the squares and thinks the meadow is beautiful if and only if

1. Not all squares are covered with grass.
2. No two mowed squares are adjacent.

Twosquares are adjacent if they share an edge. Here comes the problem: Is Tom'smeadow beautiful now?

Input

Theinput contains multiple test cases!

Eachtest case starts with a line containing two integers N, M (1<= N, M <= 10) separated by a space. Therefollows the description of Tom's Meadow. There're N lines eachconsisting of M integers separated by a space. 0(zero) meansthe corresponding position of the meadow is mowed and 1(one) means the squareis covered by grass.

A linewith N = 0 and M = 0 signals the end of theinput, which should not be processed

Output

One linefor each test case.

Output"Yes" (without quotations) if the meadow is beautiful, otherwise"No"(without quotations).

Sample Input

2 2
1 0
0 1
2 2
1 1
0 0
2 3
1 1 1
1 1 1
0 0

Sample Output

Yes
No
No

 

#include"stdio.h"int main(){      int N,M,i,j,f;   while(scanf("%d%d",&N,&M)!=EOF&&(N+M))   {int a[12][12];for(i=0;i<12;i++){for(j=0;j<12;j++){a[i][j]=1;}}f=0;for(i=1;i<=N;i++){for(j=1;j<=M;j++){scanf("%d",&a[i][j]);f+=a[i][j];}}if(f==N*M)f=0;elsef=1;for(i=1;i<=N&&f;i++){   for(j=1;j<=M&&f;j++)   {   if(!a[i][j])   {   if(a[i+1][j]+a[i][j+1]+a[i][j-1]+a[i-1][j]<4)   f=0;   }   }}printf("%s\n",f? "Yes" : "No");         }return 0;}

G - G

TimeLimit:2000MS     Memory Limit:65536KB     64bitIO Format:%lld & %llu

Submit Status Practice ZOJ 2723

Description

Prime NumberDefinition 
An integer greater than one is called a prime number if its only positivedivisors (factors) are one and itself. For instance, 2, 11, 67, 89 are primenumbers but 8, 20, 27 are not.

Semi-Prime NumberDefinition 
An integer greater than one is called a semi-prime number if it can bedecompounded to TWO prime numbers. For example, 6 is a semi-prime number but 12is not.

Your task is just todeterminate whether a given number is a semi-prime number.

Input

There are several testcases in the input. Each case contains a single integer N (2 <= N <=1,000,000)

Output

One line with a singleinteger for each case. If the number is a semi-prime number, then output"Yes", otherwise "No".

Sample Input

3
4
6
12

Sample Output

No
Yes
Yes
No

#include"stdio.h"int a[1000002];int main(){   int N;   int i,j;   for(i=2;i<=1000;i++)   {if(!a[i]){for(j=i*i;j<1000002;j+=i){a[j]=i;}}   }     while(scanf("%d",&N)!=EOF)   {      if(a[N] && !a[N/a[N]])   printf("Yes\n");   else   printf("No\n");   }return 0;}

 

H - H

Time Limit:2000MS     MemoryLimit:65536KB     64bit IO Format:%lld& %llu

Submit Status Practice ZOJ 2829

Description

Mike isvery lucky, as he has two beautiful numbers, 3 and 5. But he is so greedy thathe wants infinite beautiful numbers. So he declares that any positive numberwhich is dividable by 3 or 5 is beautiful number. Given you an integer N (1<= N <= 100000), could you please tell mike the Nth beautiful number?

Input

Theinput consists of one or more test cases. For each test case, there is a singleline containing an integer N.

Output

For eachtest case in the input, output the result on a line by itself.

Sample Input

1
2
3
4

Sample Output

3
5
6
9

 

 

#include"stdio.h"int a[100002];int main(){   int N;   int i,j=1;   for(i=3;i<=214285;i++)   {if(i%3==0||i%5==0){a[j++]=i;}   }     while(scanf("%d",&N)!=EOF)   {      printf("%d\n",a[N]);   }return 0;}