Little Boxes UVALive - 8209

题意：

给你四个数，输出四个数之和，四个数小于等于2⁶²之内

题解：

这。。。这。。水题
unsigned int 0～4294967295 (10位数，4e9)

int -2147483648～2147483647 (10位数，2e9 2^31 - 1)

long long： -9223372036854775808～9223372036854775807 (19位数， 9e18 ) 2^63 - 1

unsigned long long：0～18446744073709551615 (20位数，1e19) 2^64 - 1

代码：

#include <bits/stdc++.h>
using namespace std;
#define asd cout<<" SB "<<endl;
#define ll long long
#define INF 0x3f3f3f3f
int main(){int t;cin>>t;while(t--){ll a,b,c,d;cin>>a>>b>>c>>d;cout<<a+b+c+d<<endl;}return 0;
}

---

貌似是因为数据太水longlong才能过
大数操作用java

import java.math.BigInteger;
import java.util.Scanner;public class Main {public static Scanner s = new Scanner(System.in);public static void main(String args[]) throws Exception {Scanner in = new Scanner(System.in);int t = in.nextInt();while(t-- != 0) {BigInteger a = in.nextBigInteger();BigInteger b = in.nextBigInteger();BigInteger c = in.nextBigInteger();BigInteger d = in.nextBigInteger();System.out.println(a.add(b.add(c.add(d))));}}
}

c++模拟也可以

#include<iostream>using namespace std;
const int L=110;
string add(string a,string b){string ans;int na[L]={0},nb[L]={0};int la=a.size(),lb=b.size();for(int i=0;i<la;i++){na[la-1-i]=a[i]-'0';}for(int i=0;i<lb;i++){nb[lb-1-i]=b[i]-'0';}int lmax=la>lb?la:lb;for(int i=0;i<lmax;i++) na[i]+=nb[i],na[i+1]+=na[i]/10,na[i]%=10;if(na[lmax]) lmax++;for(int i=lmax-1;i>=0;i--) ans+=na[i]+'0';return ans;
}
string a,b,c,d;
int main(){ios::sync_with_stdio(0);cin.tie(0),cout.tie(0);int T;cin>>T;while(T--){cin>>a>>b>>c>>d;a=add(a,b);c=add(c,d);a=add(a,c);cout<<a<<endl;}return 0;
}