大数加法（主要是想记住模板）

模板可真是个好东西

题目：http://codeforces.com/gym/100735/problem/I?mobile=true

I. Yet another A + B

time limit per test

0.25 s

memory limit per test

64 MB

input

standard input

output

standard output

You are given three numbers. Is there a way to replace variables A, B and C with these numbers so the equality A + B = C is correct?

Input

There are three numbers X₁, X₂ and X₃ (1 ≤ X_i ≤ 10¹⁰⁰), each on a separate line of input.

Output

Output either "YES", if there is a way to substitute variables A, B and C with given numbers so the equality is correct, or "NO" otherwise.

Examples

Input

1
2
3

Output

YES

Input

1
2
4

Output

YES

Input

1
3
5

Output

NO

题意：从3个数中任意选3个数（可以重复选）满足a+b=c；
ps：数据特别大！（某些学了java的同学可真是让人讨厌呐）
ac代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<set>
#include<algorithm>
#include<map>
#define maxn 200005
using namespace std;
string sum(string s1,string s2)
{
    if(s1.length()<s2.length())
    {
        string temp=s1;
        s1=s2;
        s2=temp;
    }
    int i,j;
    for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--)
    {
        s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));   //注意细节
        if(s1[i]-'0'>=10)
        {
            s1[i]=char((s1[i]-'0')%10+'0');
            if(i) s1[i-1]++;
            else s1='1'+s1;
        }
    }
    return s1;
}
int main()
{
     string a[4];
     for(int i=1;i<=3;i++)
     {
         cin>>a[i];
         getchar();
     }
     int flag=0;
     for(int i=1;i<=3;i++)
     {
         for(int j=1;j<=3;j++)
         {
             for(int k=1;k<=3;k++)

             {
                if(sum(a[i],a[j])==a[k])
                    flag=1;
             }
         }
     }
     if(flag==1)cout<<"YES"<<endl;
     else cout<<"NO"<<endl;
}

然后再藏一下板子吧：

大数加法：博客链接：

http://blog.csdn.net/y990041769/article/details/19545179

string sum(string s1,string s2)
{
    if(s1.length()<s2.length())
    {
        string temp=s1;
        s1=s2;
        s2=temp;
    }
    int i,j;
    for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--)
    {
        s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));   //注意细节
        if(s1[i]-'0'>=10)
        {
            s1[i]=char((s1[i]-'0')%10+'0');
            if(i) s1[i-1]++;
            else s1='1'+s1;
        }
    }
    return s;
}

大数乘以整数型：博客链接：http://blog.csdn.net/y990041769/article/details/12645953

string Multiply(string s,int x)  //大数乘以整形数
{
    reverse(s.begin(),s.end());
    int cmp=0;
    for(int i=0;i<s.size();i++)
    {
        cmp=(s[i]-'0')*x+cmp;
        s[i]=(cmp%10+'0');
        cmp/=10;
    }
    while(cmp)
    {
        s+=(cmp%10+'0');
        cmp/=10;
    }
    reverse(s.begin(),s.end());
    return s;
}

大数除以整形数：

string Except(string s,int x)  //大数除以整形数
{
    int cmp=0,ok=0;
    string ans="";
    for(int i=0;i<s.size();i++)
    {
        cmp=(cmp*10+s[i]-'0');
        if(cmp>=x)
        {
            ok=1;
            ans+=(cmp/x+'0');
            cmp%=x;
        }
        else{
            if(ok==1)
                ans+='0';  //注意这里啊。才找出错误
        }
    }
    return ans;
}

大数乘法：http://poj.org/problem?id=2389

思想就是模拟乘法运算，用大数乘以另一个数的每一位然后大数相加就是ans

string sum(string s1,string s2)  //大数加法
{
    if(s1.length()<s2.length())
    {
        string temp=s1;
        s1=s2;
        s2=temp;
    }
    int i,j;
    for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--)
    {
        s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));   //注意细节
        if(s1[i]-'0'>=10)
        {
            s1[i]=char((s1[i]-'0')%10+'0');
            if(i) s1[i-1]++;
            else s1='1'+s1;
        }
    }
    return s1;
}

string Mult(string s,int x)  //大数乘以整形数
{
    reverse(s.begin(),s.end());
    int cmp=0;
    for(int i=0;i<s.size();i++)
    {
        cmp=(s[i]-'0')*x+cmp;
        s[i]=(cmp%10+'0');
        cmp/=10;
    }
    while(cmp)
    {
        s+=(cmp%10+'0');
        cmp/=10;
    }
    reverse(s.begin(),s.end());
    return s;
}
string Multfa(string x,string y)  //大数乘法
{
    string ans;
    for(int i=y.size()-1,j=0;i>=0;i--,j++)
    {
        string tmp=Mult(x,y[i]-'0');
        for(int k=0;k<j;k++)
            tmp+='0';
        ans=sum(ans,tmp);
    }
    return ans;
}

浮点数的n次方：博客链接：http://blog.csdn.net/y990041769/article/details/9262943

string Multiply(string s,long x)  //大数乘以整形数
{
    reverse(s.begin(),s.end());
    long cmp=0;
    for(int i=0; i<s.size(); i++)
    {
        cmp=(s[i]-'0')*x+cmp;
        s[i]=(cmp%10+'0');
        cmp/=10;
    }
    while(cmp)
    {
        s+=(cmp%10+'0');
        cmp/=10;
    }
    reverse(s.begin(),s.end());
    return s;
}
string Remove_later(string s)   //删除一个字符串的后倒0
{
    int ok=1;
    for(int i=s.size()-1; i>=0; i--)
    {
        if(s[i]=='0'){
            s.erase(i);
        }
        else if(s[i]=='.')
        {
            s.erase(i);
            ok=0;
        }
        else
            ok=0;
        if(ok==0)
            break;
    }
    return s;
}
string factorial(string s,int n)   //浮点数的n次方
{
    if(n==0)
        return "1";
    string cmp="",count="";
    long x=0,point=0;
    for(int i=0; i<s.size(); i++)
        if(s[i]!='.')
        {
            cmp+=s[i];
            x=x*10+(s[i]-'0');
        }
        else
            point=s.size()-1-i;
    for(int i=1; i<n; i++)
    {
        cmp=Multiply(cmp,x);
    }
    int ans_point=cmp.size()-n*point;
    if(ans_point<0)
    {
        count+='.';
        for(int i=ans_point; i!=0; i++)
            count+='0';
    }
    string::iterator it=cmp.begin();
    if(ans_point>=0&&ans_point<cmp.size())
        cmp.insert(it+ans_point,'.');
    count+=(Remove_later(cmp));
    return count;
}

字符串去除后导0函数，前导0可以先反转取后导。题目：http://acm.nyist.net/JudgeOnline/problem.php?pid=272

string Remove_later(string s)   //删除一个字符串的后倒0
{
    for(int i=s.size()-1; i>=0; i--)
    {
        if(s[i]=='0')
            s.erase(i);
        else
            break;
    }
    return s;
}