杭电ACM部分题答案

1000A + B Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158161 Accepted Submission(s): 50186

Problem Description

Calculate A + B.

Input

Each line will contain two integers A and B. Process to end of file.

Output

For each case, output A + B in one line.

Sample Input

1 1

Sample Output

Author

HDOJ

Statistic | Submit | Discuss | Note

#include<stdio.h>

int main()

{

int a,b;

while(scanf("%d %d",&a,&b)!=EOF)

printf("%d\n",a+b);

}

1002A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 84367 Accepted Submission(s): 15966

Problem Description

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.

Output

For each test case, you should output two lines. The first line is "Case #:", # means the n

umber of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

Sample Input

1 2

112233445566778899 998877665544332211

Sample Output

Case 1:

1 + 2 = 3

Case 2:

112233445566778899 + 998877665544332211 = 111111*********1110

Author

Ignatius.L

Statistic | Submit | Discuss | Note

#include <stdio.h>

#include <string.h>

int main(){

char str1[1001], str2[1001];

int t, i, len_str1, len_str2, len_max, num = 1, k;

scanf("%d", &t);新世纪饮食

getchar();

王冶平简历

while(t--){

int a[1001] = {0}, b[大丈夫的私房钱1001] = {0}, c[1001] = {0};

scanf("%s", str1);

len_str1 = strlen(str1);

for(i = 0; i <= len_str1 - 1; ++i)

a[i] = str1[len_str1 - 1 - i] - '0';

scanf("%s",str2);

len_str2 = strlen(str2);

for(i = 0; i <= len_str2 - 1; ++i)

b[i] = str2[len_str2 -摩擦片 1 - i] - '0';

if(len_str1 > len_str2)

len_max = len_str1;

else

len_max = len_str2;

k = 0;

for(i = 0; i <= len_max - 1; ++i){

c[i] = (a[i] + b[i] + k) % 10;

k = (a[i] + b[i] + k) / 10;

}

if(k != 0)

c[len_max] = 1;

printf("Case %d:\n", num);

num++;

printf("%s + %s = ", str1, str2);

if(c[len_max] == 1)

printf("1");

for(i = len_max - 1; i >= 0; --i){

printf("%d", c[i]);

}

printf(美能达相机维修"\n");

if(t >= 1)

printf("\n");

}

return 0;

}

1005Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44346 Accepted Submission(s): 9722

Problem Description

A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

Output

For each test case, print the value of f(n) on a single line.

Sample Input

1 1 3

1 2 10

0 0 0

Sample Output

Author

CHEN, Shunbao

Source

ZJCPC2004

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