// 面试题43：从1到n整数中1出现的次数
// 题目：输入一个整数n，求从1到n这n个整数的十进制表示中1出现的次数。例如
// 输入12，从1到12这些整数中包含1 的数字有1，10，11和12，1一共出现了5次。
#include <cstdio>
#include <cstring>
#include <cstdlib>int NumberOf1Between1AndN(int n)
{if(n<0)return 0;//数字转字符串char str[50];sprintf(str,"%d",n);//21345变为“21345”return NumberOf1(str);
}int NumberOf1(const char* str)
{if(!str||*str<'0'||*str>'9'||*str='0')return 0;//假设输入21345int first=*str-'0';//最高位字符串转整数 2unsigned int length =static_cast<unsigned int>(strlen(str));//5if(length==1&&first==0)//输入为0return 0;if(length==1&&first>0)//输入为1-9return 1;//输入str “21345”//numFirst是数字10000--19999的第一位的数目//开头是1, 后面四位为1-9里任意一个 供10×10×10×10,写一个函数实现这个计算int numFirst=0;if(first>1)//2numFirst=PowerBase10(length-1);else if(first==1)//12345numFirst=atoi(str+1)+1;//atoi为char转int， atoi(str+1)结果是2345//numOtherDigits为 1346--21345除第一位之外的数位中的数目int numOtherDigits=first*(length-1)*PowerBase10(length-1)//2*(5-1)*10*10*10//numRecursive是1---1345中的数目,递归实现int numRecursive=NumberOf1(str+1);//numberOf1("1345")return numFirst+numOtherDigits+numRecursive;}int PowerBase10(unsigned int n)
{int result=1;for(int i=0;i<n;++i)result*=10;return result;
}void Test(char* testName,int n,int expected)
{if(testName==nullptr)printf("failed n");int result=NumberOf1Between1AndN(n);if(result==expected)printf("pass n");else{printf("failed n");}}void Test()
{Test("Test1", 1, 1);Test("Test2", 5, 1);Test("Test3", 10, 2);Test("Test4", 55, 16);Test("Test5", 99, 20);Test("Test6", 10000, 4001);Test("Test7", 21345, 18821);Test("Test8", 0, 0);
}int main()
{char str[]="12345";int first=*str-'0';int num=atoi(str+1);printf("first=%d n",first);//1printf("num=%d n",num);//2345Test();return 0;
}
int NumberOf1Between1AndN(unsigned int n)
{int number = 0;for(unsigned int i = 1; i <= n; ++ i)number += NumberOf1(i);return number;
}int NumberOf1(unsigned int n)
{int number = 0;while(n){if(n % 10 == 1)number ++;n = n / 10;}return number;
}
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