Project Euler -> problem 1
2013-08-22 15:33
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1. 10以下的自然数中,属于3和5的倍数的有3,5,6和9,它们之和是23.
找出1000以下的自然数中,属于3和5的倍数的数字之和。
int main(void) {
int
i,j=0;
for(i=1;i<1000;i++){
//求3的倍数
if(i%3==0)
j+=i;
//求5的倍数
else if(i%5==0)
j+=i;
}
printf("%d",j);
return
0;
}
找出1000以下的自然数中,属于3和5的倍数的数字之和。
int main(void) {
int
i,j=0;
for(i=1;i<1000;i++){
//求3的倍数
if(i%3==0)
j+=i;
//求5的倍数
else if(i%5==0)
j+=i;
}
printf("%d",j);
return
0;
}
Answer: | 233168 |
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