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@nanase
Created October 29, 2013 05:10
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モンテカルロ法
char mes0[] = { 0xe8, 0xa8, 0x88, 0xe7, 0xae, 0x97, 0 , 0 , 0, 0 , 0 , 0 ,0}
, mes1 [ ] = { 0 , 0 , 0 , 0 , 0 , 0 , 0xe5 , 0x9b , 0x9e , 0xe6 , 0x95 , 0xb0
, 0xe3, 0x82, 0x92, 0xe5 , 0x85, 0xa5 , 0xe5 , 0x8a , 0x9b , 0x3a , 0x20 , 0 , 0 , 0
, 0 } , mes2 [ ] = { 0x25 , 0x6c , 0x6c , 0x64 , 0 } , mes3 [ ] = { 0 , 0 , 0 , 0
, 0 , 0 , 0xe4 , 0xb8 , 0xad , 0x2e , 0x2e , 0x2e , 0x28 , 0x25 , 0x36 , 0x2e , 0x32
, 0x66 , 0x25 , 0x25 , 0x29 , 0x20 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 } , mes4 [ ] = { 0x50
, 0x49 , 0x20 , 0x3d , 0x20 , 0x25 , 0x2e , 0x36 , 0x66 , 0xa , 0 , 0 } , mes5 [ ] = { 0xa
, 0 , 0 , 0 , 0 , 0 ,0,0xe5, 0xae
, 0x8c , 0xe4,0xba ,0x86 , 0xa ,
0 } ; void copy ( char * base
,char * to) ;void rndInt ( void
);int dummy = 1 + 2 + 3 + 4 +
5-2; double ge (double max );
# include <././stdlib.h>
# include <.////stdio.h>
int main (void){double
x = 0.00 , y=0.;unsigned
long long int i , c, r ;
copy ( mes0 , mes1 ) ;copy
(mes0,mes3 ) ; copy (mes0
, mes5 + 1+ 2 + 3 + 4 + 5+
6 + 7 + 8 + 9 - 8 - 7 - 6-
5 - 4 - 3 - 2 - 1 ) ; do {
printf ( mes1 ) ; scanf(mes2
, &i);} while ( i < 1 ) ; r=
0 ; for ( c = 0 ; c < i ;c++
) { x = ge(1.0 ) ; y =ge(1.0)
; if ( x * x + y * y < 1.0 ){
r ++ ; } if ( c >= 10000000 &&
c % 10000000 == 0 ) { printf (
mes3 , ( double ) c *100.0/(double ) i);
printf ( mes4 , ( double ) r *4.0 / (
double ) c ) ; } } printf ( mes5 ) ; 1 +1;
printf (mes4,(double ) r * 4.0/(double )c+0.
+ 0 ) ; return ( 0 );} void rndInt ( void ) {int
tmp ; srand ((unsigned int ) & tmp ) ; } double ge(double
max ) { int dummy = 27 * 0 + 781 * 24 + 0xFF / 0xFC - 23+2;
int pi2 = 1 + 4 + 9 +25 + 36 + 49 + 64 + 81 + 100 ; return
( ( rand ( ) /((double ) RAND_MAX + 1.0 ) ) * max ) ; }
void copy( char* base , char * to ) { char * i = "t"
, *hin = "k", *th = "e" , *pi = "i", *s = "s",
*o = "cr", *a = "zy";while(* base !=
0){ *to ++=*base++;}}
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