sum.cpp

算法实现题1-2 连续和问题 &laquo 问题描述： 给定一个正整数n

#include <string>
#include <iostream>
#include <fstream>
#include <cstdlib>
#include <cmath>
using namespace std;

int mystrcmp(const char *ch1,const char *ch2){
	int l=strlen(ch1)>strlen(ch2)?1:2;
	string str = "";
	if(l==1){
		//在第二个参数前面补零	
		int len = strlen(ch1);
		for(int k = 0;k < strlen(ch1)-strlen(ch2);k++){
			str=str+"0";
		}
		str=str+ch2;
		return strcmp(ch1,str.c_str());
	}else{
		int len = strlen(ch2);
		for(int k = 0;k < strlen(ch2)-strlen(ch1);k++){
			str=str+"0";
		}
		str=str+ch1;
		return strcmp(str.c_str(),ch2);
	}
}

string strdiv(string str,long int i){
	string tmp = "";
	char buf[2];
	int t = 0;
	int s = 0;
	char ch;
	const char * p = str.c_str();
	for(int k = 0;k < str.size();k++){
		ch = *p++;
		t = atoi(&ch);
		tmp = tmp+itoa((t+s*10)/i,buf,10);
		s = (t+s*10)%i;
	}
	//把左端的零去掉
	p = tmp.c_str();
	for(int j =0;j < tmp.size();j++){
		ch = *p++;
		if(ch!='0'){
			return &tmp[j];
		}
	}
	return "0";
}

int strmod(string str,long int i){
	int tmp = 0;
	int t = 0;
	char ch;
	const char * p = str.c_str();
	for(int k = 0;k < str.size();k++){
		ch = *p++;
		t = atoi(&ch);
		tmp = (t+tmp*10)%i;
	}
	return tmp;
}

void main(){
	ifstream in("input.txt");   
	ofstream out("output.txt");
	string str;		//接收读入的字符串		
	while(getline(in,str)){
		int count=1;	//统计满足条件的组合个数，开始时为一
		long int n = 1;		//用于后面的循环判断	
		//如果数较小，就用long int表示
		if(str.size()<10){
			const char* ch = str.c_str();
			long int linput = atol(ch);
			long int tmp = sqrt(linput/2);
			while(n<=tmp){
				if(linput%(2*n)==n){
					if(linput/(2*n)>n) count++; 
					else break;
				}
				if(linput%(2*n+1)==0){
					if(linput/(2*n+1)>n) count++;
					else break;
				}
				n++;
			}
		}
		//否则用string去处理输入的数
		else{
			string tmp = strdiv(str,2);
			char buffer[10];
			while(mystrcmp(ltoa(n,buffer,10),tmp.c_str())<=0){
				if(strmod(str,2*n)==n){
					if(mystrcmp(strdiv(str,2*n).c_str(),ltoa(n,buffer,10))>0) count++; 
					else break;
				}
				if(strmod(str,2*n+1)==0){
					if(mystrcmp(strdiv(str,2*n+1).c_str(),ltoa(n,buffer,10))>0) count++;
					else break;
				}
				n++;
			}
		}
		out<<count<<"\n";
		cout<<count<<endl;
	}
	in.close();
	out.close();
}