编程算法 - 从1到n整数中1出现的次数 代码(C)

从1到n整数中1出现的次数 代码(C)

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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.

把拆分为最高位数字, 其余数字, 最后数字求解.

21345 -> 1346-21345[10000-19999, 最高位 + 1346-x1345其余位数] + 1-1345;

代码:

/* * main.cpp * *  Created on: 2014年6月29日 *      Author: wang */#include <stdio.h>#include <stdlib.h>#include <string.h>#include <limits.h>using namespace std;int PowerBase10(size_t n) {int result = 1;for (size_t i=0; i<n; ++i)result *= 10;return result;}int NumberOf1(const char* strN) {if (!strN || *strN<'0' || *strN>'9' || *strN == '\n')return 0;int first = *strN - '0';size_t length = strlen(strN);if (length == 1 && first == 0)return 0;if (length == 1 && first > 0)return 1;//最高位数字int numFirstDight = 0;if (first > 1)numFirstDight = PowerBase10(length-1);else if (first == 1)numFirstDight = atoi(strN+1) + 1; //+1去除最高位, 在加1//其余数字int numOtherDights = first*(length-1)*PowerBase10(length-2);//最后剩余int numRecursive = NumberOf1(strN + 1);return numFirstDight + numOtherDights + numRecursive;}int NumberOf1Between1AndN (int n) {if (n<=0)return 0;char strN[50];sprintf(strN, "%d", n);return NumberOf1(strN);}int main(void){    int result = NumberOf1Between1AndN(12);    printf("result = %d\n", result);    return 0;}

输出:

result = 5