# Solution to Problem 17 on ProjectEuler.net

——-
If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.

If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of “and” when writing out numbers is in compliance with British usage.
——-

My solution:

#include

``````int main(){
int i,x,y,counter,fdigit;
int soma=0;
for (i=1;i<1001;i++){
if (i==10){
soma+=3;
continue;
}
if (i==11||i==12){
soma+=6;
continue;
}
if (i==13||i==14||i==18||i==19){
soma+=8;
continue;
}
if (i==15||i==16){
soma+=7;
continue;
}
if (i==17){
soma+=9;
continue;
}
if (i==100||i==200||i==600){
soma+=10;
continue;
}
if (i==300||i==800||i==700){
soma+=12;
continue;
}
if (i==400||i==500||i==900||i==1000){
soma+=11;
continue;
}
x=i;
counter=0;
fdigit=0;
while (x>0){
y=x%10;
if (counter==0){
if (y==1||y==2||y==6){
soma+=3;
if (y==6)
fdigit=1;
}
else if (y==4||y==5||y==9){
soma+=4;
if (y==4||y==9)
fdigit=1;
}
else if (y==3||y==8||y==7){
soma+=5;
if (y==7)
fdigit=1;
}
} /* end of if */

if (counter==1){
if (y==1){
if (fdigit==1)
soma+=4;
else
soma+=3;
}
else if (y==2||y==3||y==8||y==9)
soma+=6;
else if (y==4||y==5||y==6)
soma+=5;
else if (y==7)
soma+=7;
} /* end of if */

if (counter==2){
if (y==1||y==2||y==6)
soma+=13;
else if (y==4||y==5||y==9)
soma+=14;
else if (y==3||y==8||y==7)
soma+=15;
} /* end of if */

x = x / 10;
counter++;
} /* end of while */
} /* end of for */

printf("soma -> %dn",soma);
return 0;
}``````