A first two stages A1 = 1, A2 = 1, the later have the following relationship:
An = An-2 + 2An-1 (Note: n-2 and n-1 is a subscript)
The following procedure is prog1.c functions: turn requires M = n for an integer value corresponding to 100, 1000 and 10,000 determined so as to satisfy: Sn <M and Sn + 1> = M, where Sn = A1 + A2 + ... + An, and the n value are sequentially stored in the array element b [0], b [1 ] and b [2] Please prepare jsValue () function to achieve this function, after the function call writeDat () the array B [] in out.dat value is output to a file.
#include <stdio.h>
int b[3];
jsValue ()
{int a1=1,a2=1,a12,sn,k=2;
sn=a1+a2;
while(1)
{A12 = a1 + a2 * 2;
if(sn<100&&sn+a12>=100) b[0]=k;
if(sn<1000&&sn+a12>=1000) b[1]=k;
if(sn<10000&&sn+a12>=10000) {b[2]=k;break;}
sn = sn + a12;
a1 = a2; a2 = a12;
k++;
}
}
main()
{
jsValue ();
printf(“M=100,n=%d\nM=1000,n=%d\nM=10000,n=%d\n”,b[0],b[1],b[2]); writeDat();
}
writeDat()
{
FILE *fp;
fp=fopen(“out.dat”,“w”);
printf("%d\n%d\n%d\n",b[0],b[1],b[2]);
fprintf(fp,"%d\n%d\n%d\n",b[0],b[1],b[2]);
fclose(fp);
}
Operating results as follows:
M=100,n=6
M=1000,n=9
M=10000,n=11
