【水题】高精度加法

http://acm.hdu.edu.cn/showproblem.php?pid=1002

---

A + B Problem II
话不多说，就是很简单、很直接高精度加法。有多组用例。一般情况下都是反向字符串进行计算。

C语言实现

#include<stdio.h>#include<string.h>int T;char a[1005]={0},b[1005]={0},sum[1005]={0},tmp;int main(){    scanf("%d",&T);    int kase=0;    while ( T-- ){        scanf("%s%s",a,b);        int i,j;        printf("Case %d:\n",++kase);        printf("%s + %s = ",a,b);        for (i=0;i<1005;i++) sum[i]='0';      //清空sum为字符'0'        //反向字符串        for(j=0;a[j]!='\0';j++); j--;        for(i=0;i<j;i++,j--){            tmp=a[i];a[i]=a[j];a[j]=tmp;        }        for(j=0;b[j]!='\0';j++); j--;        for(i=0;i<j;i++,j--){            tmp=b[i];b[i]=b[j];b[j]=tmp;        }        //模拟加法        for(i=0;a[i]!='\0'&&b[i]!='\0';i++){            sum[i]+=(a[i]+b[i]-48-48);            if( sum[i]>'9' ){ //进位问题                sum[i]-=10;                sum[i+1]+=1;            }        }        if(a[i]=='\0')            for(;b[i]!='\0';i++)                sum[i]+=b[i]-48;        else            for(;a[i]!='\0';i++)                sum[i]+=a[i]-48;        for(;sum[i]!='0';i++); sum[i]='\0';  //补上最后结束符        for(j=0;sum[j]!='\0';j++);j--;//sum反串        for(i=0;i<j;i++,j--){            tmp=sum[i];sum[i]=sum[j];sum[j]=tmp;        }        printf("%s\n",sum);        if(T) printf("\n");        memset(a,0,sizeof(a));memset(b,0,sizeof(b));memset(sum,0,sizeof(sum));    }    return 0;}

C++实现

---

#include <cstdio>#include <cstring>#include <iostream>using namespace std;string add(string s1,string s2) {    int len_a=s1.length(),len_b=s2.length();    if(len_a < len_b) {        string temp = s1;        s1 = s2;        s2 = temp;    }    int i,j;    for(i = len_a-1, j = len_b-1;i>=0;i--,j--) {        s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));        if(s1[i]-'0'>=10) {            s1[i]='0'+(s1[i]-'0')%10;            if(i)   s1[i-1]++;            else    s1='1'+s1;        }    }    return s1;}string A,B,C;int main(){    while(cin>> A>> B) {        C=add(A,B);        cout<< C<< endl;    }    return 0;}