B Split a Number （字符求和 ）
Dima worked all day and wrote down on a long paper strip his favorite number nnconsisting of ll digits. Unfortunately, the strip turned out to be so long that it didn't fit in the Dima's bookshelf.
To solve the issue, Dima decided to split the strip into two non-empty parts so that each of them contains a positive integer without leading zeros. After that he will compute the sum of the two integers and write it down on a new strip.
Dima wants the resulting integer to be as small as possible, because it increases the chances that the sum will fit it in the bookshelf. Help Dima decide what is the minimum sum he can obtain.
Input
The first line contains a single integer ll (2≤l≤1000002≤l≤100000) — the length of the Dima's favorite number.
The second line contains the positive integer nn initially written on the strip: the Dima's favorite number.
The integer nn consists of exactly ll digits and it does not contain leading zeros. Dima guarantees, that there is at least one valid way to split the strip.
Output
Print a single integer — the smallest number Dima can obtain.
Examples
7
1234567
1801
3
101
11
Note
In the first example Dima can split the number 12345671234567 into integers 12341234 and 567567. Their sum is 18011801.
In the second example Dima can split the number 101101 into integers 1010 and 11. Their sum is 1111. Note that it is impossible to split the strip into "1" and "01" since the numbers can't start with zeros.
这个能过CF但是有BUG，但能 hack 死一大片（下面的样例），正确解法是记录非0位置，二分查计算
6
000034
会输出0034，很明显错误
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
//
#define lson rt<<1, l, m
#define rson rt<<1|1, m+1, r
#define gcd __gcd
#define mem(s,t) memset(s,t,sizeof(s))
#define debug(a) for(int i=1;i<=n;i++) cout<<" "<<a[i]; cout<<endl;
#define debug(a) for(int i=0;i<n;i++) {for(int j=0;j<m;j++) cout<<a[i][j]<<" "; cout<<endl; }
#define read1(a) for (int i = 1; i <= (int)n; i++) {cin>>a[i]); }
#define read2(a) for(int i=0;i<n;i++) {for(int j=0;j<m;j++) cin>>a[i][j] ; }
#define TLE std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cout.precision(10);
#define fi first
#define se second
#define pb push_back
#define pql priority_queue<lli>
#define pq priority_queue<int>
#define ok return 0;
#define oi(x) cout<<x<<endl;
#define os(str) cout<<string(str)<<endl;
using namespace std;
inline void NO()
{
cout<<"NO\n";
}
inline void YES()
{
cout<<"YES\n";
}
typedef long long lli;
typedef long double ld;
const int N = 1e5+5;
const int logN = (int)log2(N)+1;
const ld PI = acos(-1);
const ld EPS = 1e-10;
typedef pair < int, int > pii;
typedef pair < lli, lli > pll;
typedef vector < lli > vl;
typedef vector < int > vi;
multiset <int> ms;
#define rep(ttt,k) for (int i=ttt;i<=k;i++)
#define per(ttt,k) for (int i=ttt;i>=k;i--)
const int mxn = 2e5+10;
int a[mxn];
/*
#ifdef LOCAL
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
*/
string sum(string s1 , string s2)
{
if(s1.size()<s2.size())
swap(s1,s2);
for(int i=0 ; i<s2.size();i++)
s1[ s1.size()-s2.size()+i ] += s2[i]-'0';
for( int i=s1.size()-1;i>=1;i--)
{
if(s1[i]>'9')
{
s1[i] -= 10;
s1[i-1]++;
}
}
if(s1[0]>'9')
{
s1[0]-=10;
s1 = '1'+s1;
}
//cout<<s1<<endl;
return s1;
}
string fun(string s1 , string s2)
{
if(s1.size()<s2.size()) return s1;
if(s1.size()>s2.size()) return s2;
return s1<s2?s1:s2;
}
int main()
{
int n; string str,ch;
cin>>n>>str;
ch = str;
int mid = n>>1;
while(str[mid]=='0') mid--;
if( str[mid]!='0' && mid>=0)
ch = fun( ch , sum( str.substr(0,mid) , str.substr(mid) ) );
mid = (n+1)>>1;
while(mid<n && str[mid]=='0')
mid++;
if( str[mid]!='0' && mid<n )
ch = fun( ch , sum( str.substr(0,mid) , str.substr(mid) ) );
cout<<ch<<endl;
}
#include <bits/stdc++.h> #define fi first #define se second const int N = 100100; using namespace std; int n; char a[N]; string get(int x) { vector < int > A, B; for(int i = x; i > 0; i--){ A.push_back(a[i] - '0'); } for(int i = n; i > x; i--){ B.push_back(a[i] - '0'); } while(A.size() < B.size()) A.push_back(0); while(B.size() < A.size()) B.push_back(0); A.push_back(0); for(int i = 0; i < B.size(); i++){ A[i] += B[i]; A[i + 1] += A[i] / 10; A[i] %= 10; } while(A.back() == 0) A.pop_back(); string res = ""; while(!A.empty()){ res += char(A.back() + '0'); A.pop_back(); } return res; } int main() { //freopen("input.txt", "r", stdin); //freopen("output.txt", "w", stdout); ios_base::sync_with_stdio(0); cin >> n; string res = ""; vector < int > v; for(int i = 1; i <= n; i++){ cin >> a[i]; res += "9"; if(i < n){ v.push_back(i); } } sort(v.begin(), v.end(), [&](int x, int y){ return abs(x - n / 2) > abs(y - n / 2); }); for(int i = 0; i < 10; i++){ while(!v.empty() && a[v.back() + 1] == '0'){ v.pop_back(); } if(v.empty()){ break; } string can = get(v.back()); v.pop_back(); if(can.size() < res.size() || can.size() == res.size() && can < res){ res = can; } } cout << res << "\n"; }
#pragma GCC optimize("O3") #include <bits/stdc++.h> using namespace std; //defines typedef long long ll; typedef long double ld; #define TIME clock() * 1.0 / CLOCKS_PER_SEC #define prev _prev #define y0 y00 //permanent constants const ld pi = acos(-1.0); const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int digarr[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6}; const int dx[4] = {0, 1, 0, -1}; const int dy[4] = {1, 0, -1, 0}; const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0}; const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1}; const int alf = 26; const int dig = 10; const int two = 2; const int th = 3; const ll prost = 239; const ll bt = 30; const ld eps = 1e-9; const ll INF = (ll)(1e18 + 239); const int BIG = (int)(2e9 + 239); const int MOD = 1e9 + 7; //998244353; const ll MOD2 = (ll)MOD * (ll)MOD; //random mt19937_64 rnd(239); //(chrono::high_resolution_clock::now().time_since_epoch().count()); //constants const int M = (int)(2e5 + 239); const int N = (int)(2e3 + 239); const int L = 20; const int T = (1 << 18); const int B = (int)sqrt(M); const int X = 1e4 + 239; string sum(string s, string t) { vector<int> ans(max((int)s.size(), (int)t.size()) + 1, 0); for (int i = (int)s.size() - 1; i >= 0; i--) ans[(int)s.size() - 1 - i] += s[i] - '0'; for (int i = (int)t.size() - 1; i >= 0; i--) ans[(int)t.size() - 1 - i] += t[i] - '0'; int p = 0; for (int i = 0; i < ans.size(); i++) { int ost = (p + ans[i]) % 10; p = (p + ans[i]) / 10; ans[i] = ost; } while (ans.back() == 0) ans.pop_back(); string u = ""; while (!ans.empty()) { u += (char)('0' + ans.back()); ans.pop_back(); } return u; } int n; string s; string gett(int i) { string s1 = ""; for (int x = 0; x < i; x++) s1 += s[x]; string s2 = ""; for (int x = i; x < n; x++) s2 += s[x]; return sum(s1, s2); } bool cmp(string a, string b) { if ((int)a.size() != (int)b.size()) return (int)a.size() < (int)b.size(); return (a < b); } int32_t main() { #ifdef ONPC freopen("input.txt", "r", stdin); #endif ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); cin >> n >> s; vector<int> cand; int k = 0; for (int i = n / 2; i < n && k < 3; i++) if (s[i] != '0') { cand.push_back(i); k++; } k = 0; for (int i = n / 2; i >= 1 && k < 3; i--) if (s[i] != '0') { cand.push_back(i); k++; } vector<string> v; for (int i : cand) v.push_back(gett(i)); sort(v.begin(), v.end(), cmp); cout << v[0]; return 0; }
