資源簡介
給定一整數n,計算所有小于等于n的非負整數中數字1出現的個數
代碼片段和文件信息
#include?
#include?
int?getNums(int?num);
int?getEveryRes(int?start?int?end?int?len);
int?main()?{
????int?num?=?-1;
????static?int?count1?=?0;
????int?length?=?-1;
????//?判斷當前輸入數是幾位數
????printf(“input?a?num?:?“);
????scanf(“%d“?&num);
????length?=?getNums(num);
????printf(“你輸入的是:%d?位數\n“?length);
????for?(int?i?=?1;?i?<=?length;?++i)??//?位數?1?1-9??2??10?-99?3?100-999
????{
????????if?(i?==?length)
????????{
????????????count1?+=?getEveryRes(pow(10?i?-?1)?num?i);
????????}
????????else
????????{
????????????count1?+=?getEveryRes(pow(10?i?-?1)?pow(10?i)?-?1?i);
????????}
????}
????printf(“%d?-?%d?中1出現的個數?=?%d\n“?1?n
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