POJ 1053 Integer Inquiry && HDOJ 1047 Integer Inquiry (大数加法)

题目链接(POJ):http://poj.org/problem?id=1503

题目链接(HDOJ):http://acm.hdu.edu.cn/showproblem.php?pid=1047

Integer Inquiry

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 30856		Accepted: 12007

Description

One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking varIoUs sums of those numbers. 
``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were here to see these results.'' (Chip moved to a new apartment,once one became available on the third floor of the Lemon Sky apartments on Third Street.) 

Input

The input will consist of at most 100 lines of text,each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length,and will only contain digits (no VeryLongInteger will be negative). 

The final input line will contain a single zero on a line by itself. 

Output

Your program should output the sum of the VeryLongIntegers given in the input.

Sample Input

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

Sample Output

370370367037037036703703703670

Source

East Central North America 1996

题意:求n个大数之和

题解; 大数加法模板题——POJ和HDU上格式出处的要求稍有不同——HDOJ上的描述好混乱。

AC代码(POJ)

#include<iostream>
#include<cstring>
#include<string>
#define maxn 300
using namespace std;
int numx[maxn],numy[maxn],n;
string str,tmp;
string Add(string x,string y){
    string res="";
    memset(numx,sizeof(numx));
    memset(numy,sizeof(numy));
    int lenx=x.size(),leny=y.size();
    int maxlen=lenx>leny ? lenx:leny;
    for(int i=0;i<lenx;i++)numx[lenx-i-1]=x[i]-'0';
    for(int i=0;i<leny;i++)numy[leny-i-1]=y[i]-'0';
    for(int i=0;i<=maxlen;i++){
        numx[i]+=numy[i];
        if(numx[i]>9){
            numx[i+1]+=numx[i]/10;
            numx[i]%=10;
        }
    }
    int i=maxlen+2;
    for(;i>0&&!numx[i];)i--;
    for(;i>=0;i--)res+=numx[i]+'0';
    return res;
}
int main()
{
    string sum="0";
    while(cin>>str&&str!="0")
    sum=Add(sum,str);
    cout<<sum<<endl;
    return 0;
}

AC代码(HDU):

#include<iostream>
#include<cstring>
#include<string>
#define maxn 300
using namespace std;
int numx[maxn],leny=y.size();
    int maxlen=lenx>leny ? lenx:leny;
    for(int i=0;i<lenx;i++)numx[lenx-i-1]=x[i]-'0';
    for(int i=0;i<leny;i++)numy[leny-i-1]=y[i]-'0';
    for(int i=0;i<=maxlen;i++){
        numx[i]+=numy[i];
        if(numx[i]>9){
            numx[i+1]+=numx[i]/10;
            numx[i]%=10;
        }
    }
    int i=maxlen+2;
    for(;i>0&&!numx[i];)i--;
    for(;i>=0;i--)res+=numx[i]+'0';
    return res;
}
int main()
{
    while(cin>>n){
        while(n--){
                string sum="0";
            while(cin>>str&&str!="0"){
                sum=Add(sum,str);
            }
            cout<<sum<<endl;
            if(n)cout<<endl;
        }
    }
    return 0;
}

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作者:MummyDing

出处:http://www.voidcn.com/article/p-oghcrafq-bbd.html