Some water questions in the school competition, water questions in the school Competition

1236: xq and Crystal MineTime Limit: 1 Sec Memory Limit: 128 MB

Submit: 313 Solved: 76

[Submit] [Status] [Web Board] Description

Xq once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n> 1) is an n struct * contains n matrix with a diamond inscribed into it.

You are given an odd integer n. you need to draw a crystal of size n. the diamond cells of the matrix shocould be represented by character "D ". all other cells of the matrix shocould be represented by character "*". look at the examples to understand what you need to draw.

Input

The only line contains an integer n (3 <= n <= 101; n is odd ).

Output

Output a crystal of size n.

Sample Input

357

Sample Output

*D*DDD*D***D***DDD*DDDDD*DDD***D*****D*****DDD***DDDDD*DDDDDDD*DDDDD***DDD*****D***

HINT

Source

Q

**AC code:**

#include <cstdio>#include <cstring>#include <algorithm>using namespace std; char a[105][105]; int main(){ int n; while(scanf("%d", &n)!=EOF) { for(int i=0; i<n; i++) { for(int j=0; j<n; j++) { a[i][j] = '*'; } } for(int i=0; i<=n/2; i++) { for(int j=n/2-i; j<=n/2+i; j++) { a[i][j] = 'D'; } } for(int i=n/2; i<n; i++) { for(int j=n/2-(n-i-1); j<=n/2+(n-i-1); j++) { a[i][j] = 'D'; } } for(int i=0; i<n; i++) { for(int j=0; j<n; j++) { printf("%c", a[i][j]); } printf("\n"); } } return 0;}

1246: xq problem Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 351 Solved: 119

[Submit] [Status] [Web Board] Description

Xq has known that Gauss has used a formula to calculate the positive integers and numbers between 1 and 100. Therefore, he has come up with many similar problems, first, demonstrate your worship of Gauss, and second, show how smart you are. Each time xq comes up with a question that can be quickly answered by itself. Gradually, xq has a lingering power and is very simple at the beginning. You don't have to think about it to get the answer. Later on, it takes a long time to answer the question correctly. Now xq has encountered the biggest problem for myself. This question has been answered by xq for three days and three nights. When xq was about to give up the answer, xq encountered a computer. xq heard that the computing speed of the computer was very fast, and xq was about to try again. Of course, he asks you to help him solve this difficult problem. The problem of xq is similar to that of Gaussian. The difference is that xq wants to know the sum of all the even numbers from 1 to n (including 1 and n.

Input

There are multiple groups of test data. You need to process the data to the end of the file (EOF ).

Each group of data contains a positive integer n (1 <=n <= 100 ). Test data are separated by line breaks.

Output

The sum of all the even numbers from 1 to n is output for each group of test data.

Sample Input

13

Sample Output

02

HINT

Source

Q

AC code:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std; int main(){ int n; while(scanf("%d", &n)!=EOF) { int ans = 0; for(int i=1; i<=n; i++) { if(i%2==0)ans+=i; } printf("%d\n", ans); } return 0;}

1240: Text HolesTime Limit: 1 Sec Memory Limit: 128 MB

Submit: 281 Solved: 94

[Submit] [Status] [Web Board] Description

Xq wrote some text on a piece of paper and now he wants to know how many holes are in the text. What is a hole? If you think of the paper as the plane and a letter as a curve on the plane, then each letter divides the plane into regions. for example letters "A", "D", "O", "P", "Q" "R ", divide the plane into two regions so we say these letters each have one hole. similarly, letter "B" has two holes and letters such as "C", "E", "F", "K" have no holes. we say that the number of holes in the text is equal to the total number of holes in the letters of the text. help Chef to determine how many holes are in the text.

Input

The first line contains a single integer T <= 40, the number of test cases. T test cases follow. the only line of each test case contains a non-empty text composed only of uppercase letters of English alphabet. the length of the text is less then 100. there are no any spaces in the input.

Output

For each test case, output a single line containing the number of holes in the corresponding text.

Sample Input

2CODECHEFDRINKEATCODE

Sample Output

25

HINT

Source

Q

AC code:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std; char a[110]; int judge(char a){ if(a == 'A')return 1; else if(a == 'D') return 1; else if(a == 'O') return 1; else if(a == 'P') return 1; else if(a == 'Q') return 1; else if(a == 'R') return 1; else if(a == 'B') return 2; else return 0;} int main(){ int T; scanf("%d", &T); while(T--) { scanf("%s", a); int ans = 0, len = strlen(a); for(int i=0; i<len; i++) { ans +=judge(a[i]); } printf("%d\n", ans); } return 0;}

1241: Fibonacci number Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 590 Solved: 68

[Submit] [Status] [Web Board] Description

Xq has recently become infatuated with the Fibonacci series, because when n tends to be infinite, the ratio of the next item to the previous item is getting closer and closer to the Golden division of 0.618. But what makes xq unhappy is that the Fibonacci number is growing too fast, so that when n is large, the 32-bit integer or even 64-bit integer in the computer cannot represent this number. For this reason, xq has encountered this problem. xq wants to know whether the number of the K (0 <= K <= 10 ^ 10000) is an odd or even number, because K is too big, xq does not know how to solve this problem. Now xq wants you to solve this problem. The Fibonacci series is the following series: 0, 1, 2, 3, 5 ......, Its formula is F (n) = F (n-1) + F (n-2), (n> = 2), where F (0) = 0, F (1) = 1.

Input

The first line, T, indicates that there are 1 <= T <100 test samples.

In the next T row, each row has one data K (0 <= K <= 10 ^ 10000), which indicates the item to be determined.

Output

If the K-th item is an even number, the output is YES; otherwise, the output is NO.

Sample Input

201

Sample Output

YESNO

HINT

Source

Q

AC code:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std; char a[10005]; int main(){ int T; scanf("%d", &T); while(T--) { scanf("%s", a); int m = 0, len = strlen(a); for(int i=0; i<len; i++) { m += a[i]-'0'; } if(m%3==0)printf("YES\n"); else printf("NO\n"); } return 0;}

1242: Tang xuecang's honeymoon sequel Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 288 Solved: 15

[Submit] [Status] [Web Board] Description

In October 1 this year, Tang xuecang went out on a tour with his girlfriend again! This time Tang took his girlfriend to the beautiful city of north China South-xinyang.

Tang xuecang, who accompanied his girlfriend in the Pedestrian Street shopping mall, was surprised to find that Xiaoqiang, a junior high school math teacher, is now switching to a snack bar. As a math teacher of Tang xuecang, Xiao Qiang wanted to know the current math level of Tang xuecang, so he gave Tang xuecang a difficult number theory problem, and said to Tang, "if you can solve this problem, I will treat you tonight! ".

Now we know two numbers, gcd and lcm. If Tang xuecang has two numbers, a and B (a <= B), the maximum common divisor is gcd, and the minimum common factor is lcm.

Input

**The input contains multiple groups of samples!**In each group, enter an integer T (T <120), followed by T rows. Each row has two numbers gcd and lcm (1 <= gcd <= lcm <2 ^ 31)

Output

Corresponding to the input in each line. If a B does not meet the condition, the output is "senior Tang is so smart !", Otherwise, a B is output (a and B are separated by spaces). If multiple groups of solutions exist, the output corresponds to the group with the smallest a value,**For details about the output format, see the example.**

Sample Input

21 23 4

Sample Output

1 2senior Tang is so smart!

HINT

Assume that you have obtained the GCD function (the maximum common number.

Evaluate the LCM (least public multiple) function as follows:

Int lcm (int a, int B ){

Return a/gcd (a, B) * B;

}

Not complete...

Source

Yin Hua

AC code:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std; int main(){ int gcd, lcm, T; while(scanf("%d", &T)!=EOF) { while(T--) { scanf("%d %d", &gcd, &lcm); if(gcd<=lcm && lcm%gcd==0) printf("%d %d\n", gcd, lcm); else printf("senior Tang is so smart!\n"); } } return 0;}

**Link: Official warm-up match**