poj1000-1002

1000

num 1000

#include <iostream>

using namespace std;

int main()
{
	int a,b;
	cin >> a >> b;
	cout << a+b << endl;
	return 0;
}

 poj 1001
  version:1.0
  author:Knight
  Email:S.Knight.Work@gmail.com


#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<memory.h>
using namespace std;
 
char Result[200];//存R^N的结果
 
//大实数的乘法,乘数为FirMultiplier和SecMultiplier，结果存在Result中
void HigRealMul(char* FirMultiplier, char* SecMultiplier, char* Result);
//剔除实数尾部的无效0或小数点
void CutInsignificantTail(char* StrR);
//计算小数点在实数中的位数
int CountPointIndex(char* StrR);
//删除实数中的小数点,PointIndex为小数点在实数中从右向左数的第几位
void DeletePoint(char* StrR, int PointIndex);
 
int main(void)
{
    char StrR[10];//R对应的字符串
    int N;
    int i;
    int PointIndex = 0;//记录小数点在实数中从右向左数的第几位，如数字1.26在第3位，数字4的小数点在第0位
 
    while(scanf("%s%d", StrR, &N) != EOF)
    {
        memset(Result, 0, 200);
 
        CutInsignificantTail(StrR);
 
        PointIndex = CountPointIndex(StrR);//compute the dot position
 
        DeletePoint(StrR, PointIndex);
 
        strcpy(Result, StrR);
 
        for (i=2; i<=N; i++)
        {
            HigRealMul(Result, StrR, Result);
        }
 
        int Len = strlen(Result);
 
        if (Len -(PointIndex - 1) * N < 0)
        {
            printf(".");
            for (i = Len - (PointIndex - 1) * N; i<0; i++)
            {
                printf("0");
            }
        }
 
        for (i=0; i<Len; i++)
        {
            //输出小数点
            if (i == Len -(PointIndex - 1) * N)
            {
                printf(".");
            }
            printf("%c", Result[i]);
        }
        printf("\n");
        //printf("%s\n", Result);
        //printf("%d\n", PointIndex);
    }
    return 0;
}
 
//大实数的乘法,乘数为FirMultiplier和SecMultiplier，结果存在Result中
void HigRealMul(char* FirMultiplier, char* SecMultiplier, char* Result)
{
 
    char TmpResult[200];
    int i,j;
    int k = -1;//控制TmpResult[]下标
    int FirLen = strlen(FirMultiplier);
    int SecLen = strlen(SecMultiplier);
 
    memset(TmpResult, '0', 200);
 
//模拟乘法运算
    for (i=SecLen-1; i>=0; i--)
    {
        k++;
 
        int FirMul;
        int SecMul = SecMultiplier[i] - '0';
        int Carry;//进位
 
        for (j=FirLen-1; j>=0; j--)
        {
            FirMul = FirMultiplier[j] - '0';
            TmpResult[k + FirLen - 1 - j] +=   FirMul * SecMul % 10;
            Carry = FirMul * SecMul / 10 + (TmpResult[k + FirLen - 1 - j] - '0') / 10;
            TmpResult[k + FirLen - 1 - j] = (TmpResult[k + FirLen - 1 - j] - '0') % 10 + '0';
            TmpResult[k + FirLen - j] += Carry;
        }
    }
 
//防止某一位的值超过9
    for (k=0; k<200; k++)
    {
        TmpResult[k + 1] += (TmpResult[k] - '0') / 10;
        TmpResult[k] = (TmpResult[k] - '0') % 10 + '0';
    }
//将设置字符串结束符
    for (k=199; k>=0; k--)
    {
        if ('0' != TmpResult[k - 1])
        {
            TmpResult[k] = '\0';
            break;
        }
    }
 
//将临时存储的答案TmpResult倒转变成我们熟悉的方式，存到Result中
    for (i=strlen(TmpResult)-1,j=0; i>=0; i--,j++)
    {
        Result[j] = TmpResult[i];
    }
    Result[j] = '\0';
 
}
 
//剔除实数尾部的无效0或小数点
void CutInsignificantTail(char* StrR)
{
    int i;
    int PointIndex = CountPointIndex(StrR);
    int Len = strlen(StrR);
 
    if (0 == PointIndex)
    {
        if ('.' == StrR[Len - 1])
        {
            StrR[Len - 1] = '\0';
        }
 
        return;
    }
 
    for (i=Len-1; i>Len-1-PointIndex; i--)
    {
        if ('0' == StrR[i] || '.' == StrR[i])
        {
            StrR[i] = '\0';
        }
        else
        {
            return ;
        }
    }
}
 
//计算小数点在实数中的位数
int CountPointIndex(char* StrR)
{
    int i;
    int Index = 0;
 
    for (i = strlen(StrR); i>=0; i--)
    {
 
        if ('.' == StrR[i])
        {
            break;
        }
        else
        {
            Index++;
        }
    }
 
    if (-1 == i)
    {
        Index = 0;
    }
 
    return Index;
 
}
 
//删除实数中的小数点
void DeletePoint(char* StrR, int PointIndex)
{
    int i;
    int Len = strlen(StrR);
 
    for (i=strlen(StrR)-PointIndex; i<Len; i++)
    {
        StrR[i] = StrR[i+1];
    }
}

1002 电话号码的去重和统计

#include<iostream>
#include<string>
#include<stdlib.h>
#include<cstdio>
#include<cstring>
using namespace std;
int numint[110000]={0};
char number[110000][100]={0};
int compare(const void *a, const void *b)
{
	return(*(int *)a-*(int *)b);
}
int main()
{
	int n=0,i=0,j=0,N=0,point=1000000,d[40]={2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,9,9,9,9},k=0;
	cin>>n;
	getchar();
	for(i=0;i<n;i++)
	{
		gets(number[i]);
		N=strlen(number[i]);
		point=1000000;
		for(j=0;j<N;j++)
		{
			if(number[i][j]=='-')
			{
				continue;
			}
			if(number[i][j]<='9'&&number[i][j]>='0')
			{
				numint[i]+=(number[i][j]-'0')*point;
				point=point/10;
				continue;
			}
			numint[i]+=d[number[i][j]-'A']*point;
			point=point/10;
		}
	}
	qsort(numint,n,sizeof(int),compare);
	for(i=0;i<n;i++)
	{
		for(j=i+1;j<n;j++)
		{
			if(numint[i]!=numint[j])
			{
				if(j-i>1)
				{
					printf("%03d-%04d %d\n", numint[i]/10000, numint[i]%10000, j-i);
					i=j-1;
					k=1;
					break;
				}
				else
				{
					i=j-1;
					break;
				}
			}
		}
		if(numint[i]==numint[j-1]&&j==n&&i!=j-1)
		{
			printf("%03d-%04d %d\n", numint[i]/10000, numint[i]%10000, j-i);
			k=1;
			break;
		}
	}
	if(!k)
	{
		cout<<"No duplicates."<<endl;
	}

}