hdu1002
题意：大数相加
#include<bits/stdc++.h> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <string> #include <stdio.h> #include <queue> #include <stack> #include <map> #include <set> #include <string.h> #include <vector> typedef long long ll ; #define int ll #define mod 1000000007 #define gcd __gcd #define rep(i , j , n) for(int i = j ; i <= n ; i++) #define red(i , n , j) for(int i = n ; i >= j ; i--) #define ME(x , y) memset(x , y , sizeof(x)) //ll lcm(ll a , ll b){return a*b/gcd(a,b);} ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;} //int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;} //const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len} #define SC scanf #define INF 0x3f3f3f3f #define PI acos(-1) #define pii pair<int,int> #define fi first #define se second #define pb push_back #define mp make_pair #define all(v) v.begin(),v.end() #define size(v) (int)(v.size()) #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 using namespace std; const int N = 1e6+7; const int maxn = 5e2+9; int cnt ; string plu(string a , string b){//大数加最优版本 int i , j ; if(size(a) < size(b)) swap(a , b);//a长度更长 for(i = size(a)-1 , j = size(b)-1 ; i >= 0 ; i-- , j--){//不逆置，直接从后双指针模拟 a[i] = a[i] + (j >=0 ? b[j] - '0' : 0) ; if(a[i]-'0' >= 10){ a[i] = a[i] - 10 ; if(i) a[i-1]++; else a = '1' + a ;//答案只可能比a长度长1且为1. } } return a; } void solve(){ if(cnt) cout << endl; string a , b ; cin >> a >> b; cout << "Case " << ++cnt << ":" << endl; cout << a << " + " << b << " = " << plu(a , b) << endl; } signed main() { //ios_base::sync_with_stdio(false); //cin.tie(NULL);cout.tie(NULL); int _ ;cin>>_;while(_--) solve(); }