標籤:mystra 編程演算法 1出現的次數 代碼 c
從1到n整數中1出現的次數 代碼(C)
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題目: 輸入一個整數n, 求從1到n這n個整數的十進位表示中1出現的次數.
把拆分為最高位元字, 其餘數字, 最後數字求解.
21345 -> 1346-21345[10000-19999, 最高位 + 1346-x1345其餘位元] + 1-1345;
代碼:
/* * 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;//最高位元字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//其餘數字int numOtherDights = first*(length-1)*PowerBase10(length-2);//最後剩餘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;}
輸出:
result = 5