输入两只非负10上前制整数A和B(&lt输入两独非负10前进制整数A和B(&lt

B1022.
D进制的A+B (20)

B1022.
D进制的A+B (20)

Description:

Description:

输入两单非负10上制整数A和B(<=230-1),输出A+B的D
(1 < D <= 10)进制数。

输入两独非负10迈入制整数A和B(<=230-1),输出A+B的D
(1 < D <= 10)进制数。

Input:

Input:

输入在一行中相继被出3单整数A、B和D。

输入在一行中逐条被有3个整数A、B和D。

Output:

Output:

输出A+B的D进制数。

输出A+B的D进制数。

Sample
Input:

Sample
Input:

123 456
8

123 456
8

Sample
Output:

Sample
Output:

1103

1103

 1 #include <cstdio>
 2 
 3 int main()
 4 {
 5     int a, b, d;
 6     scanf("%d%d%d", &a, &b, &d);
 7 
 8     int sum = a+b;
 9     int ans[31], num = 0;
10     do {
11         ans[num++] = sum%d;
12         sum /= d;
13     } while(sum != 0);
14 
15     for(int i=num-1; i>=0; --i)
16         printf("%d", ans[i]);
17 
18     return 0;
19 }
 1 #include <cstdio>
 2 
 3 int main()
 4 {
 5     int a, b, d;
 6     scanf("%d%d%d", &a, &b, &d);
 7 
 8     int sum = a+b;
 9     int ans[31], num = 0;
10     do {
11         ans[num++] = sum%d;
12         sum /= d;
13     } while(sum != 0);
14 
15     for(int i=num-1; i>=0; --i)
16         printf("%d", ans[i]);
17 
18     return 0;
19 }

 

 

A1019.
General Palindromic Number (20)

A1019.
General Palindromic Number (20)

Description:

Description:

A number
that will be the same when it is written forwards or backwards is known
as a Palindromic Number. For example, 1234321 is a palindromic number.
All single digit numbers are palindromic numbers.

A number
that will be the same when it is written forwards or backwards is known
as a Palindromic Number. For example, 1234321 is a palindromic number.
All single digit numbers are palindromic numbers.

Although
palindromic numbers are most often considered in the decimal system, the
concept of palindromicity can be applied to the natural numbers in any
numeral system. Consider a number N > 0 in base b >= 2, where it
is written in standard notation with k+1 digits ai as the sum
of (aibi) for i from 0 to k. Here, as usual, 0
<= ai < b for all i and ak is non-zero. Then
N is palindromic if and only if ai = ak-i for all
i. Zero is written 0 in any base and is also palindromic by
definition.

Although
palindromic numbers are most often considered in the decimal system, the
concept of palindromicity can be applied to the natural numbers in any
numeral system. Consider a number N > 0 in base b >= 2, where it
is written in standard notation with k+1 digits ai as the sum
of (aibi) for i from 0 to k. Here, as usual, 0
<= ai < b for all i and ak is non-zero. Then
N is palindromic if and only if ai = ak-i for all
i. Zero is written 0 in any base and is also palindromic by
definition.

Given any
non-negative decimal integer N and a base b, you are supposed to tell if
N is a palindromic number in base b.

Given any
non-negative decimal integer N and a base b, you are supposed to tell if
N is a palindromic number in base b.

Input:

Input:

Each input
file contains one test case. Each case consists of two non-negative
numbers N and b, where 0 <= N <= 109 is the decimal
number and 2 <= b <= 109 is the base. The numbers are
separated by a space.

Each input
file contains one test case. Each case consists of two non-negative
numbers N and b, where 0 <= N <= 109 is the decimal
number and 2 <= b <= 109 is the base. The numbers are
separated by a space.

Output:

Output:

For each
test case, first print in one line “Yes” if N is a palindromic number in
base b, or “No” if not. Then in the next line, print N as the number in
base b in the form “ak ak-1 … a0“.
Notice that there must be no extra space at the end of output.

For each
test case, first print in one line “Yes” if N is a palindromic number in
base b, or “No” if not. Then in the next line, print N as the number in
base b in the form “ak ak-1 … a0“.
Notice that there must be no extra space at the end of output.

Sample
Input1:

Sample
Input1:

27 2

27 2

Sample
Output1:

Sample
Output1:

Yes
1 1 0 1
1

Yes
1 1 0 1
1

Sample
Input2:

Sample
Input2:

121 5

121 5

Sample
Output2:

Sample
Output2:

No

No

4 4 1

4 4 1

 1 #include <cstdio>
 2 
 3 #define MaxSize 50
 4 int ans[MaxSize];
 5 
 6 int main()
 7 {
 8     //freopen("E:\\Temp\\input.txt", "r", stdin);
 9 
10     int N, b, num = 0;
11     bool flag = true;
12     scanf("%d %d", &N, &b);
13 
14     do {
15         ans[num++] = N%b;
16         N /= b;
17     } while(N != 0);
18 
19     for(int i=0, j=num-1; i<=(num-1)/2, j>=(num-1)/2; ++i, --j) {
20         if(ans[i] != ans[j]) {
21             flag = false;
22             break;
23         }
24     }
25 
26     if(flag == true)    printf("Yes\n");
27     else printf("No\n");
28     for(int i=num-1; i>=0; --i) {
29         printf("%d", ans[i]);
30         if(i != 0)  printf(" ");
31     }
32 
33     return 0;
34 }

 1 #include <cstdio>
 2 
 3 bool judge(int z[], int num)
 4 {
 5     for(int i=0; i<=num/2; ++i) {
 6         if(z[i] != z[num-1-i])  return false;
 7         else return true;
 8     }
 9 }
10 
11 int main()
12 {
13     int n, b, z[40], num = 0;
14     scanf("%d%d", &n, &b);
15 
16     do {
17         z[num++] = n%b;
18         n /= b;
19     } while(n != 0);
20     bool flag = judge(z, num);
21 
22     if(flag == true)    printf("Yes\n");
23     else printf("No\n");
24     for(int i=num-1; i>=0; --i) {
25         printf("%d", z[i]);
26         if(i != 0)  printf(" ");
27     }
28 
29     return 0;
30 }
 1 #include <cstdio>
 2 
 3 #define MaxSize 50
 4 int ans[MaxSize];
 5 
 6 int main()
 7 {
 8     //freopen("E:\\Temp\\input.txt", "r", stdin);
 9 
10     int N, b, num = 0;
11     bool flag = true;
12     scanf("%d %d", &N, &b);
13 
14     do {
15         ans[num++] = N%b;
16         N /= b;
17     } while(N != 0);
18 
19     for(int i=0, j=num-1; i<=(num-1)/2, j>=(num-1)/2; ++i, --j) {
20         if(ans[i] != ans[j]) {
21             flag = false;
22             break;
23         }
24     }
25 
26     if(flag == true)    printf("Yes\n");
27     else printf("No\n");
28     for(int i=num-1; i>=0; --i) {
29         printf("%d", ans[i]);
30         if(i != 0)  printf(" ");
31     }
32 
33     return 0;
34 }

 1 #include <cstdio>
 2 
 3 bool judge(int z[], int num)
 4 {
 5     for(int i=0; i<=num/2; ++i) {
 6         if(z[i] != z[num-1-i])  return false;
 7         else return true;
 8     }
 9 }
10 
11 int main()
12 {
13     int n, b, z[40], num = 0;
14     scanf("%d%d", &n, &b);
15 
16     do {
17         z[num++] = n%b;
18         n /= b;
19     } while(n != 0);
20     bool flag = judge(z, num);
21 
22     if(flag == true)    printf("Yes\n");
23     else printf("No\n");
24     for(int i=num-1; i>=0; --i) {
25         printf("%d", z[i]);
26         if(i != 0)  printf(" ");
27     }
28 
29     return 0;
30 }

 

 

A1027.
Colors in Mars (20)

A1027.
Colors in Mars (20)

Description:

Description:

People in
Mars represent the colors in their computers in a similar way as the
Earth people. That is, a color is represented by a 6-digit number, where
the first 2 digits are for Red, the middle 2 digits for Green, and the
last 2 digits for Blue. The only difference is that they use radix 13
(0-9 and A-C) instead of 16. Now given a color in three decimal numbers
(each between 0 and 168), you are supposed to output their Mars RGB
values.

People in
Mars represent the colors in their computers in a similar way as the
Earth people. That is, a color is represented by a 6-digit number, where
the first 2 digits are for Red, the middle 2 digits for Green, and the
last 2 digits for Blue. The only difference is that they use radix 13
(0-9 and A-C) instead of 16. Now given a color in three decimal numbers
(each between 0 and 168), you are supposed to output their Mars RGB
values.

Input:

Input:

Each input
file contains one test case which occupies a line containing the three
decimal color values.

Each input
file contains one test case which occupies a line containing the three
decimal color values.

Output:

Output:

For each
test case you should output the Mars RGB value in the following format:
first output “#”, then followed by a 6-digit number where all the
English characters must be upper-cased. If a single color is only
1-digit long, you must print a “0” to the left.

For each
test case you should output the Mars RGB value in the following format:
first output “#”, then followed by a 6-digit number where all the
English characters must be upper-cased. If a single color is only
1-digit long, you must print a “0” to the left.

Sample
Input:

Sample
Input:

15 43
71

15 43
71

Sample
Output:

Sample
Output:

#123456

#123456

 1 #include <cstdio>
 2 
 3 char radix[13] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C'};
 4 
 5 int main()
 6 {
 7     int r, g, b;
 8     scanf("%d%d%d", &r, &g, &b);
 9 
10     printf("#");
11     printf("%c%c", radix[r/13], radix[r%13]);
12     printf("%c%c", radix[g/13], radix[g%13]);
13     printf("%c%c", radix[b/13], radix[b%13]);
14 
15     return 0;
16 }
 1 #include <cstdio>
 2 
 3 char radix[13] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C'};
 4 
 5 int main()
 6 {
 7     int r, g, b;
 8     scanf("%d%d%d", &r, &g, &b);
 9 
10     printf("#");
11     printf("%c%c", radix[r/13], radix[r%13]);
12     printf("%c%c", radix[g/13], radix[g%13]);
13     printf("%c%c", radix[b/13], radix[b%13]);
14 
15     return 0;
16 }

 

 

A1058. A+B
in Hogwarts (20)

A1058. A+B
in Hogwarts (20)

Description:

Description:

If you are a
fan of Harry Potter, you would know the world of magic has its own
currency system — as Hagrid explained it to Harry, “Seventeen silver
Sickles to a Galleon and twenty-nine Knuts to a Sickle, it’s easy
enough.” Your job is to write a program to compute A+B where A and B are
given in the standard form of “Galleon.Sickle.Knut” (Galleon is an
integer in [0, 107], Sickle is an integer in [0, 17), and
Knut is an integer in [0, 29)).

If you are a
fan of Harry Potter, you would know the world of magic has its own
currency system — as Hagrid explained it to Harry, “Seventeen silver
Sickles to a Galleon and twenty-nine Knuts to a Sickle, it’s easy
enough.” Your job is to write a program to compute A+B where A and B are
given in the standard form of “Galleon.Sickle.Knut” (Galleon is an
integer in [0, 107], Sickle is an integer in [0, 17), and
Knut is an integer in [0, 29)).

Input:

Input:

Each input
file contains one test case which occupies a line with A and B in the
standard form, separated by one space.

Each input
file contains one test case which occupies a line with A and B in the
standard form, separated by one space.

Output:

Output:

For each
test case you should output the sum of A and B in one line, with the
same format as the input.

For each
test case you should output the sum of A and B in one line, with the
same format as the input.

Sample
Input:

Sample
Input:

3.2.1
10.16.27

3.2.1
10.16.27

Sample
Output:

Sample
Output:

14.1.28

14.1.28

 1 #include <cstdio>
 2 
 3 int main()
 4 {
 5     int a[3], b[3], c[3];
 6     scanf("%d.%d.%d %d.%d.%d", &a[0], &a[1], &a[2], &b[0], &b[1], &b[2]);
 7 
 8     int carry = 0;
 9     c[2] = (a[2]+b[2])%29;
10     carry = (a[2]+b[2])/29;
11     c[1] = (a[1]+b[1]+carry)%17;
12     carry = (a[1]+b[1]+carry)/17;
13     c[0] = a[0]+b[0]+carry;
14 
15     printf("%d.%d.%d", c[0], c[1], c[2]);
16 
17     return 0;
18 }
 1 #include <cstdio>
 2 
 3 int main()
 4 {
 5     int a[3], b[3], c[3];
 6     scanf("%d.%d.%d %d.%d.%d", &a[0], &a[1], &a[2], &b[0], &b[1], &b[2]);
 7 
 8     int carry = 0;
 9     c[2] = (a[2]+b[2])%29;
10     carry = (a[2]+b[2])/29;
11     c[1] = (a[1]+b[1]+carry)%17;
12     carry = (a[1]+b[1]+carry)/17;
13     c[0] = a[0]+b[0]+carry;
14 
15     printf("%d.%d.%d", c[0], c[1], c[2]);
16 
17     return 0;
18 }