Novice small white, find a big with me-CodePudding

Programming to solve 1! + 2! + 3! +... + 19! + 20! Value, the problem how to do it, I should be in the process of, the result has been operating errors,

CodePudding user response:

20! A lot, to hold, for reference:

 # include 

Int main () 
{
int i; 
__int64 s=0, TMP=1; 
for(i=1; i<=20; I++) 
{
TMP *=I; 
S +=TMP; 
} 
Printf (" % I64d \ n ", s); 

return 0; 
}

CodePudding user response:

If this is clear some

 int main () 
{
/* 
* 1! + 2! + 3! +... + 19! + 20! 
* 
* 1 * 1 + 2 + 3 * 2 * 1 * 1 + 4 * 3 * 2 * 1 + 
* */

//a number is used to store each factorial 
Double last=1; 
//a number used to sum the 
double sum=0; 
//from 1 to 20 
For (int I=1; I & lt;=20; I++) 
{
The last *=I;//I=1=& gt; (1) I=1 * 2=& gt; (2 * 1 * 1) I=3=& gt; 3 * (2 * 1 * 1) I=4=& gt; 4 * (3 * 2 * 1 * 1) 
The sum +=last;//added together to 
} 
//print the results 
Printf (" lf the sum=% \ n ", sum); 
return 0; 
}

CodePudding user response:

Try just now, thank you,
But in the fourth row and the second line of code, I don't understand the

CodePudding user response:

refer to the second floor small white ah really reply:

this is understand some

 int main () 
{
/* 
* 1! + 2! + 3! +... + 19! + 20! 
* 
* 1 * 1 + 2 + 3 * 2 * 1 * 1 + 4 * 3 * 2 * 1 + 
* */

//a number is used to store each factorial 
Double last=1; 
//a number used to sum the 
double sum=0; 
//from 1 to 20 
For (int I=1; I & lt;=20; I++) 
{
The last *=I;//I=1=& gt; (1) I=1 * 2=& gt; (2 * 1 * 1) I=3=& gt; 3 * (2 * 1 * 1) I=4=& gt; 4 * (3 * 2 * 1 * 1) 
The sum +=last;//added together to 
} 
//print the results 
Printf (" lf the sum=% \ n ", sum); 
return 0; 
}

Got it, thank you,

CodePudding user response:

reference qq_45773596 reply: 3/f

try just now, thank you,
But in the fourth row and the second line of code, I don't understand the

When doing ACM problem, often will meet some of the bigger of the two integers, and the commonly used built-in integer types often appear too small: long and int scope
Is [- 2 ^ 31, 2 ^ 31), namely - 2147483648 ~ 2147483647, and unsigned range is [0, 2 ^ 32), namely, 0 ~ 4294967295, that is,
Regular 32-bit integer only able to handle the following number 4 billion,
The encounter is bigger than 4 billion number? At this time will use c + + 64 - bit extensions, different compilers for the expansion of the 64 - bit integer
Different, VC6.0 64 - bit integer respectively called __int64 and unsigned __int64, its scope is [- 2 ^ 63, 2 ^ 63), and [0, 2 ^ 64), namely - 9223372036854775808 ~ 9223372036854775807 and 0 ~ 18446744073709551615 (180 billion), for a 64 - bit integer arithmetic and 32-bit integer basic same, support arithmetic and bit operations,

CodePudding user response:

refer to the second floor small white ah really reply:

this is understand some

 int main () 
{
/* 
* 1! + 2! + 3! +... + 19! + 20! 
* 
* 1 * 1 + 2 + 3 * 2 * 1 * 1 + 4 * 3 * 2 * 1 + 
* */

//a number is used to store each factorial 
Double last=1; 
//a number used to sum the 
double sum=0; 
//from 1 to 20 
For (int I=1; I & lt;=20; I++) 
{
The last *=I;//I=1=& gt; (1) I=1 * 2=& gt; (2 * 1 * 1) I=3=& gt; 3 * (2 * 1 * 1) I=4=& gt; 4 * (3 * 2 * 1 * 1) 
The sum +=last;//added together to 
} 
//print the results 
Printf (" lf the sum=% \ n ", sum); 
return 0; 
}

Test for a moment, your program output result:
Sum=2561327494111820288.000000
The individual feels this result is wrong, because this result should be an odd number, rather than an even number;

The last and the sum is defined as an unsigned long long after the result is
Sum=2561327494111820313

Will be the last and the sum is defined as a long double was no problem, after received:
Sum=2561327494111820313.000000

In addition, for a long double output. Suggest use % 0 lf format output, because the class and will not be a floating point number (real), was supposed to get an integer;

CodePudding user response:

reference 6 building self-confidence boy reply:

Quote: refer to the second floor small white ah really reply:
this is understand some

int main ()
{
/*
* 1! + 2! + 3! +... + 19! + 20!
*
* 1 * 1 + 2 + 3 * 2 * 1 * 1 + 4 * 3 * 2 * 1 +
* */

//a number is used to store each factorial
Double last=1;
//a number used to sum the
double sum=0;
//from 1 to 20
For (int I=1; I & lt;=20; I++)
{
The last *=I;//I=1=& gt; (1) I=1 * 2=& gt; (2 * 1 * 1) I=3=& gt; 3 * (2 * 1 * 1) I=4=& gt; 4 * (3 * 2 * 1 * 1)
The sum +=last;//added together to
}
//print the results
Printf (" lf the sum=% \ n ", sum);
return 0;
}

Test for a moment, your program output result:
Sum=2561327494111820288.000000
The individual feels this result is wrong, because this result should be an odd number, rather than an even number;

The last and the sum is defined as an unsigned long long after the result is
Sum=2561327494111820313

Will be the last and the sum is defined as a long double was no problem, after received:
Sum=2561327494111820313.000000

In addition, for a long double output. Suggest use % 0 lf format output, because the class and will not be a floating point number (real), was supposed to get an integer;

I'm dizzy, which is a method for ah

CodePudding user response:

reference 6 building self-confidence boy reply:

Quote: refer to the second floor small white ah really reply:

this is understand some

 int main () 
{
/* 
* 1! + 2! + 3! +... + 19! + 20! 
* 
* 1 * 1 + 2 + 3 * 2 * 1 * 1 + 4 * 3 * 2 * 1 + 
* */

//a number is used to store each factorial 
Double last=1; 
//a number used to sum the 
double sum=0; 
//from 1 to 20 
For (int I=1; I & lt; nullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnull