(完整版)杭电acm部分答案

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

#include

void main()

{

int a,b;

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

{

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

}

}

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.

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

#include

void main()

{

int n,sum,i;

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

{

sum=0;

for( i=0;i<=n;i++)

sum+=i;

printf("%d\n\n",sum);

}

}

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 number 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

2

1 2

112233445566778899 998877665544332211

Sample Output

Case 1:

1 +

2 = 3

Case 2:

112233445566778899 + 998877665544332211 = 1111111111111111110 #include

#include

int main(){

char str1[1001], str2[1001];

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

scanf("%d", &t);

getchar();

while(t--){

int a[1001] = {0}, b[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 - 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("\n");

if(t >= 1)

printf("\n");

}

return 0;

}