// The sum of two large integers
#include <stdio.h>
#include <stdlib.h>
#define N 100
void printArrayInt(int *arr, int n, int ncol){
int i;
for (i=0; i<n; i++){
printf("%d ",arr[i]);
if(!((i+1)%ncol)) printf("\n");
}
printf("\n");
}
void printArrayInverseInt(int *arr, int n, int ncol){
int i;
for (i=0; i<n; i++){
printf("%d ",arr[n-1-i]);
if(!((i+1)%ncol)) printf("\n");
}
printf("\n");
}
int add(int *a, int *b, int curr_len, int s){
//add b to a
int i,temp,temp1=0;
for(i=0;i<curr_len;i++){
temp = a[i+s]+b[i]+temp1;
a[i+s]=temp%10;
temp1 = (int)(temp/10);
}
if(temp1>0){
a[s+curr_len]=temp1;
curr_len++;
}
return curr_len+s;
}
int main(){
int i,len,a[N+1], b[N];
srand(1);
for (i=0;i<N;i++){
a[i] = rand() % 10;
b[i] = rand() % 10;
}
while(!a[N-1]) a[N-1] = rand() % 10;
while(!b[N-1]) b[N-1] = rand() % 10;
printArrayInverseInt(a,N,25);
printArrayInverseInt(b,N,25);
len = add(a,b,N,0);
printArrayInverseInt(a,len,25);
printf("length of a is %d\n",len);
return 0;
}2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
More generally, what is the smallest positive number that is evenly divisible by all of the numbers from 1 to N? Here N could be large, such as 10000.
Refer to the Section “Loops continue and arrays” and “Arrays”.
#include<stdio.h>
#include<math.h>
#define N 100
#define M 1000
int isPrime(int n){
int i=2;
double sqrtn=sqrt(1.0*n);
while(i<=sqrtn){
if(!(n%i)) return 0;
i++;
}
return 1;
}
int product(int fact[], int n, int curr_len, int digits){
int i,temp1=0;
curr_len+=digits-1;
for(i=0;i<curr_len;i++){
temp1=fact[i]*n+temp1/10;
fact[i]=temp1%10;
}
while(temp1/=10){
fact[i]=temp1;
curr_len++;
}
return curr_len;
}
int isDivededBy1toN(int fact[],int n){
int i,j,m0;
int curr_len=1,digits,m;
for(i=2;i<=n;i++)
if(isPrime(i)){
m0=j=i;
while((j*=i)<=n) m0 = j;
digits = 1;
m=m0;
while(m/=10) digits++;
curr_len = product(fact,m0,curr_len,digits);
printf("j=%d\n",m0);
}
return curr_len;
}
int main(){
int i,fact[M],curr_len;
for(i=0;i<M;i++) fact[i]=0;
fact[0]=1;
curr_len = isDivededBy1toN(fact,N);
for(i=curr_len-1;i>=0;i--) printf("%d",fact[i]);
return 0;
}// A magic square is a square grid (where n is the number of cells on each side)
// filled with distinct positive integers in the range.
// such that each cell contains a different integer
// and the sum of the integers in each row, column and diagonal is equal,
// where the sum equals (n^2+1)*n/2.
#include<stdio.h>
#define n 11
//n must be odd.
int main(){
int i,l,k,flag,half;
int a[n][n],mid;
half=(n-1)/2;
mid=(n*n+1)/2;
a[half][half]=mid;
for(i=1;i<half+1;i++){
for(l=0;l<2;l++){
flag=1;
if(l%2==1) flag=-1;
a[half-i*flag][half] = mid+(2*i*(i+1)-2*i+2)*flag;
a[half][half-i*flag] = mid+(2*i*(i+1)-2*i)*flag;
a[half-i*flag][half-i*flag] = mid-(2*i*(i+1)-2*i+1)*flag;
a[half-i*flag][half+i*flag] = mid-(2*i*(i+1)-2*i-1)*flag;
}
for(l=1;l<i;l++){
for(k=0;k<2;k++){
flag=1;
if(k%2==1) flag=-1;
a[half-i*flag][half-i+l] = mid+(2*i*(i+1)-(l-1)*2)*flag;
a[half-i*flag][half+l] = mid-(2*i*(i+1)-2*i-2-(l-1)*2)*flag;
a[half-i+l][half-i*flag] = mid-(2*i*(i+1)-1-(l-1)*2)*flag;
a[half+l][half-i*flag] = mid+(2*i*(i+1)-2*i-3-(l-1)*2)*flag;
}
}
}
printf("The a[%d][%d] are :\n\n",n,n);
for(i=0;i<n;i++){
for(k=0;k<n;k++)
printf("%6d",a[i][k]);
printf("\n");
}
return 0;
}// Calculate the first 50 pairs (p,q) of Pell equation.
#include<stdio.h>
void pf2(int *a1, int *b1, int *Lab, int N){
int i,La,ta,tb,ta1,tb1,n=1;
a1[0]=3;
b1[0]=2;
Lab[0]=Lab[1]=1;
while(n<N){
ta=tb=0;La=Lab[0];
for(i=0;i<La;i++){
ta1=a1[i]*3+b1[i]*4+ta;
tb1=a1[i]*2+b1[i]*3+tb;
a1[i]=ta1%10;
b1[i]=tb1%10;
ta=ta1/10;
tb=tb1/10;
}
Lab[0]=La;
Lab[1]=La;
if(ta){
a1[i]=ta;
Lab[0]++;
}
if(tb){
b1[i]=tb;
Lab[1]++;
}
//printf("n=%d \t La=%d \t Lb = %d\n",n, Lab[0], Lab[1]);
n++;
}
}
int main(){
int i;
int a[100], b[100], Lab[2];
for (i = 0; i < 100; i++) a[i]=b[i]=0;
pf2(a,b,Lab,50);
printf("p=");
for(i=Lab[0]-1;i>=0;i--)
printf("%d",a[i]);
printf(" q=");
for(i=Lab[1]-1;i>=0;i--)
printf("%d",b[i]);
printf("\n");
return 0;
}