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C试题及答案(国外英语资料)
C++试题及答案(国外英语资料)
1. Title: polynomial 1! +2! +3! +...... +15! Sum of values
Stored in variable out
#include iostream.h
Void, main ()
{
Int, out, I, J, factorial;
Out = 0;
For (I = 1; I 15; i++) {
Factorial = 1;
For (J = 1; J I; j++) factorial = factorial * j;
Out = out + factorial;
}
Cout out endl;
}
2. Title: seek all prime numbers between 1 and 200, and put the prime number into the variable sum
#include iostream.h
Int prime (int i)
{
Int j;
For (J = 2; J I; j++) {
If (... (I% J)) return 0;
}
If (J = = I) return j;
Else return 0;
}
Void, main ()
{
Int, I, sum;
Sum = 0;
For (I = 1; I 200; i++) {
Sum = sum + prime (I);
}
Cout sum endl;
}
3. Title: using while loop programming, natural numbers between 1 to 100 odd squares
And sum.
#include iostream.h
Void, main ()
{
Int, I, sum;
I = 1;
Sum = 0;
While (I = 100) {
Sum = (I * I);
I = I + 2;
}
Cout sum endl;
}
4. Title: to determine whether a number 23437 is a prime number
The parameter flag, flag is 1, stands for prime number, 0 represents no)
#include iostream.h
Int prime (int i)
{
Int j;
For (J = 2; J I; j++)
{
If (... (I% J)) return 0;
}
If (J = = I) return j;
Else return 0;
}
Void, main ()
{
Int, I, flag;
I = 23437;
Flag = prime (I) \i;
Cout flag endl;
}
5. Title: known a number m (=252), ask you sum of numbers
#include iostream.h
#include string
Using namespace std;
Void, main ()
{
String s;
Int, len, sum, i;
S = 252;
Len = s.size ();
Sum = 0;
For (I = 0; I len; i++)
Sum = sum + s[i] -0; / / take a digital character, it is the digital value minus the character0
Cout, the sum of your numbers is: sum endl;
}
6. Title: the odd number between 1-100 cumulative order into the N, until its and so on
At or greater than 200
#include iostream.h
Void, main ()
{
Int, I, sum;
I = 1;
Sum = 0;
While (I = 100) {
Sum = sum + i;
I = I + 2;
If (sum = 200 break);
}
Cout sum endl;
}
7. Title: two numbers to calculate x with the division method, the largest convention y
number
#includeiostream.h
Void, main ()
{
Int, a, B, num1,
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