[1228] Bored Three-God

心已赠人 2022-02-26 10:54 140阅读 0赞

*  ### \[1228\] Bored Three-God ###
 *  时间限制: 1000 ms　内存限制: 65535 K
 *  问题描述
 *  The bored Three-God get another boring question.
    
    This problem is ask you plus two big nubmer, please help him, when you solve this problem you can
    
    speak to Three-God,"Oh, Please give me a diffculte one, this one is too easy".
 *  输入
 *  There are muti-case  
    For each case, there are two integers, n, m (0 < n, m < 10^10000).
 *  输出
 *  Calculate two integers' sum
 *  样例输入
 *      1 1
        2 3
        10000 100000
 *  样例输出
 *      2
        5
        110000
 *  提示
 *      无
 *  来源
 *      Three-God
 *  操作

## 题目大意：大数加法！！ ##

## 思路：超级简单，但是要注意如果输入有前导0，则输出也要有前导0（说实话，第一次见到这么恶心的大数，还带前导0的） ##

代码：

#include<map>
    #include<set>
    #include<stack>
    #include<queue>
    #include<cmath>
    #include<string>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define ll unsigned long long
    #define inf 0x3f3f3f3
    #define esp 1e-8
    #define bug {printf("mmp\n");}
    #define mm(a,b) memset(a,b,sizeof(a))
    #define T() int test,q=1;scanf("%d",&test); while(test--)
    const int maxn=2e6+100;
    const double pi=acos(-1.0);
    const int N=2e6;
    const int mod=1e9+7;
    char s1[N],s2[N];
    int a[N],b[N],c[N];
    void add()
    {
        mm(a,0);
        mm(b,0);
        mm(c,0);
        int l1=strlen(s1);
        int l2=strlen(s2);
        int lm=max(l1,l2);
        int la=0,lb=0;
        for(int i=l1-1;i>=0;i--)
            a[la++]=s1[i]-'0';
        for(int i=l2-1;i>=0;i--)
            b[lb++]=s2[i]-'0';
        for(int i=0;i<lm;i++)
        {
            c[i]+=a[i]+b[i];
            c[i+1]+=c[i]/10;
            c[i]%=10;
        }
        if(c[lm])
            printf("%d",c[lm]);
        for(int i=lm-1;i>=0;i--)
           printf("%d",c[i]);
        printf("\n");
    }
    
    int main()
    {
        while(scanf("%s %s",s1,s2)!=EOF)
        {
            add();
        }
        return 0;
    }