description
Known: Sn = 1 + 1/2 + 1/3 + ... + 1 / n. Clearly for any integer K, when n is large enough, Sn is larger than K.
Are now given an integer K (1 <= k <= 15), requires a minimum of n-calculated; such that Sn> K.
Enter an integer K. Output an integer n.
Wrong answers and analysis:
#include <cstdio>
the using namespace STD ; int main () { int I , K ; a float n- = 0.0 , Sn = 0 ; // here noted that the data range should not use double type Scanf ( "% D" , & K ); the while (Sn <= K ) {n- + = . 1 ; Sn + = 1.0 / n- ; } the printf ( "% D" , ( int ) n- ); return 0 ; } attached:
| Types of | Place | range |
|---|---|---|
| char | 1 byte | -128 to 127 or 0 to 255 |
| unsigned char | 1 byte | 0 to 255 |
| signed char | 1 byte | -128 to 127 |
| int | 4 bytes | -2147483648 to 2147483647 |
| unsigned int | 4 bytes | 0-4294967295 |
| signed int | 4 bytes | -2147483648 to 2147483647 |
| short int | 2 bytes | -32768 to 32767 |
| unsigned short int | 2 bytes | 0 to 65,535 |
| signed short int | 2 bytes | -32768 to 32767 |
| long int | 8 bytes | -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 |
| signed long int | 8 bytes | -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 |
| unsigned long int | 8 bytes | 0 to 18,446,744,073,709,551,615 |
| float | 4 bytes | +/- 3.4e +/- 38 (~ 7 digits) |
| double | 8 bytes | +/- 1.7e +/- 308 (~ 15 digits) |
| long double | 16 bytes | +/- 1.7e +/- 308 (~ 15 digits) |
| wchar_t | 2 or 4 bytes | A wide character |
Correct answer:
#include<cstdio>
using namespace std;
int main() { int i,k; double n=0.0,sn=0; scanf("%d",&k); while(sn<=k) { n+=1; sn+=1.0/n; } printf("%d",(int)n); return 0; }