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send + more = money C solution
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#include <stdio.h> | |
#define LENGTH(array)( (int)(sizeof(array)/sizeof(array[0])) ) | |
/* | |
problem : | |
s e n d | |
+ m o r e | |
--------- | |
m o n e y | |
mapping: | |
s e n d m o r y | |
0 1 2 3 4 5 6 7 | |
*/ | |
int LINE1[] = { 0, 1, 2, 3 }; | |
int LINE2[] = { 4, 5, 6, 1 }; | |
int RESULT[] = { 4, 5, 2, 1, 7 }; | |
int values[] = { 0, 1, 2, 3, 4, 5, 6, 7 }; | |
int test_proposal() { | |
int number1 = 0, number2 = 0, result = 0, i, l; | |
/* We refuse number starting with 0 */ | |
if ((values[LINE1[0]] == 0) || (values[LINE2[0]] == 0) || (values[RESULT[0]] == 0)) { | |
return 0; | |
} | |
for(i=0,l=LENGTH(LINE1); i<l; ++i) number1 = number1*10 + values[LINE1[i]]; | |
for(i=0,l=LENGTH(LINE2); i<l; ++i) number2 = number2*10 + values[LINE2[i]]; | |
for(i=0,l=LENGTH(RESULT);i<l; ++i) result = result*10 + values[RESULT[i]]; | |
i = (number1 + number2) == result; | |
if(i){ | |
printf("Solution found: \n %4d \n+ %4d \n= %5d", number1, number2, result); | |
} | |
return i; | |
} | |
int free_digits[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }; | |
int free_digits_count = 10; | |
int make_proposal(int varID) { | |
int i, tmp; | |
for (i = 0; i < free_digits_count; ++i) { | |
tmp = free_digits[i]; | |
values[varID] = tmp; | |
if(varID+1 < LENGTH(values)) { | |
//Swap, decrease | |
free_digits_count--; | |
free_digits[i] = free_digits[free_digits_count]; | |
free_digits[free_digits_count] = tmp; | |
if(make_proposal(varID+1)) return 1; | |
//Swap, increase | |
free_digits[free_digits_count] = free_digits[i]; | |
free_digits[i] = tmp; | |
free_digits_count++; | |
}else{ | |
if(test_proposal()) return 1; | |
} | |
} | |
return 0; | |
} | |
int main() { | |
int found = make_proposal(0); | |
if (!found){ | |
printf("No solution found:\n"); | |
} | |
} |
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