求斐波那契数列第N项及大数求和

/******************************************************************************** * 版权所有(c) YYYY, <JJYY>, 保留所有权利 *******************************************************************************/ #include "stdafx.h"#include <iostream.h>#include <string.h>#define MAXLEN 2000#define MAX(a, b) (((a)>(b))?(a):(b))#define OUTchar* add(char* pStr1, char* pStr2);char* FacBigFun(int n, OUT char* Total);int main(int argc, char* argv[]){char Total[MAXLEN] = {0};cout << FacBigFun(2009, Total) << endl;return 0;cout << add("823", "823") << endl;return 0;}/*! @function********************************************************************************函数名 : FacFun功能 : 用循环求斐波那契数列第N项参数 : n项返回值 : 第n项的结果用int表示抛出异常 : 无*******************************************************************************/ /*int FacFun(int n){//特殊情况if ((n == 1) || (n == 2)){return 1;}int nOne = 1;int nTwo = 1;int nTotal = 0;for (int i = 3; i <= n; i++){nTotal = nOne + nTwo;nOne = nTwo;nTwo = nTotal;}return nTotal;}*//*! @function********************************************************************************函数名 : FacFun功能 : 重载大数之和求循环求斐波那契数列第N项(n比较大的时)参数 : n项返回值 : 第n项的结果用char*表示抛出异常 : 无*******************************************************************************/ char* FacBigFun(int n , OUT char* Total){//特殊情况if ((n == 1) || (n == 2)){return "1";}char One[MAXLEN] = "1";char Two[MAXLEN] = "1";memset(Total, 0, MAXLEN);for (int i = 3; i <= n; i++){memcpy(Total, add(One, Two), MAXLEN);//nTotal = nOne + nTwo;memcpy(One, Two, MAXLEN);//nOne = nTwo;memcpy(Two, Total, MAXLEN);//nTwo = nTotal;cout << i << " " << Total << endl;}return Total;}/*! @function********************************************************************************函数名 : add功能 : 求两个大数之和参数 : 两个用字符串表示的大数返回值 : 大数之和抛出异常 : 无*******************************************************************************/ char* add(char* pStr1, char* pStr2){int nLen1 = strlen(pStr1);int nLen2 = strlen(pStr2);int maxLen = MAX(nLen1, nLen2);int nStr1 = 0;int nStr2 = 0;int nFlag = 0;char Total[MAXLEN] = {0};for (int i = 0; i < maxLen; i ++){if (nLen1-i>=1){nStr1 = pStr1[nLen1-i-1] - '0';}else{nStr1 = 0;}if (nLen2-i>=1){nStr2 = pStr2[nLen2-i-1] - '0';}else{nStr2 = 0;}Total[i] = (nStr1 + nStr2 + nFlag)%10 + '0';nFlag = (nFlag + nStr1 + nStr2)/10;}if (nFlag == 1){Total[i] = '1';}int nLen = strlen(Total);char Result[MAXLEN] = {0};for (int j = 0; j < nLen; j++){Result[j] = Total[nLen-j-1];}return Result;}