Jim

### 3 problems has been added

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 Solved problems from projecteuler.net ... ...
problem1/prog 0 → 100755
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problem1/prog.c 0 → 100644
 #include "stdio.h" #define PRECIS 1000 int multiple (int n){ int i; int result; int summ; result = 0; summ =0; i = 1; while (result < PRECIS){ summ += result; printf ("Result :%i\t Summ:%i\n",result, summ); result = n*i; i++; } return summ; } int multiple_check (int n, int m){ int i; int result; int summ; result = 0; summ =0; i = 1; while (result < PRECIS){ if ( (result % m)) { summ += result; printf ("Result :%i\t Summ:%i\n",result, summ); } else { printf ("Do nothing! :)\n"); } result = n*i; i++; } return summ; } main () { int total; int total1; int total2; total1 = multiple(3); total2 = multiple_check(5,3); total = total1 + total2; printf ("Total1: %i Total2: %i Total: %i\n ", total1, total2, total); return 0; } ... ...
problem2/prog2 0 → 100755
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problem2/prog2.c 0 → 100644
 #include "stdio.h" #define STEP 4000000 main () { long int sum; int one, two, three; one=1; two=2; three = 0; sum = 0; while (two <= STEP){ three = one + two; // printf ("%i,", three); if ( (two % 2 ) == 0) sum += two; // printf ("sum is now: %lu\n", sum); one = two; two = three; } printf ("Sum = %lu\n", sum); return 0; } ... ...
problem3/proc3.c 0 → 100644
 #include #include #define NUMBER 600851475143 void primefactors(long int n){ long int i; while (n % 2 == 0){ printf ("1 %i\n" , 2); n = n/2; } for (i=3; i <= sqrt(n); i= i+2){ while (n % i == 0){ printf("2 %li\n", i); n = n/i; } } if (n > 2) printf ("3 %li\n", n); } main () { primefactors (NUMBER); return 0; } ... ...
problem3/prog3 0 → 100755
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problem3/prog3.c 0 → 100644
 #include #include #define NUMBER 600851475143 void primefactors(long int n){ long int i; while (n % 2 == 0){ printf ("1 %i\n" , 2); n = n/2; } for (i=3; i <= sqrt(n); i= i+2){ while (n % i == 0){ printf("2 %li\n", i); n = n/i; } } if (n > 2) printf ("3 %li\n", n); } main () { primefactors (NUMBER); return 0; } ... ...