1000 A + B Problem
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
2
Author
HDOJ
代码：
#include
int a,b;
while(scanf(\,&a,&b)!=EOF) printf(\,a+b); }
1001 Sum Problem
Problem Description
Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).
In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.
1
Input
The input will consist of a series of integers n, one integer per line.
Output
For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.
Sample Input
1 100
Sample Output
1 5050
Author
DOOM III
解答：
#include
int n,i,sum; sum=0;
while((scanf(\,&n)!=-1)) {
sum=0;
for(i=0;i<=n;i++)
sum+=i;
printf(\,sum); } }
2
1002 A + B Problem II
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 \the test case. The second line is the an equation \Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2 1 2
112233445566778899 998877665544332211
Sample Output
Case 1: 1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
Author
Ignatius.L
代码：
#include
int main(){
char str1[1001], str2[1001];
int t, i, len_str1, len_str2, len_max, num = 1, k; scanf(\, &t); getchar(); while(t--){
3
int a[1001] = {0}, b[1001] = {0}, c[1001] = {0}; scanf(\, str1);
len_str1 = strlen(str1);
for(i = 0; i <= len_str1 - 1; ++i)
a[i] = str1[len_str1 - 1 - i] - '0'; scanf(\,str2);
len_str2 = strlen(str2);
for(i = 0; i <= len_str2 - 1; ++i)
b[i] = str2[len_str2 - 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(\, num); num++;
printf(\, str1, str2); if(c[len_max] == 1) printf(\);
for(i = len_max - 1; i >= 0; --i){ printf(\, c[i]); }
printf(\); if(t >= 1)
printf(\); }
return 0; }
1005 Number Sequence
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).
4