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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#521907#5460. Sum of Numbersno_RED_no_DEADTL 0ms3604kbC++202.6kb2024-08-16 16:35:232024-08-16 16:35:23

Judging History

你现在查看的是最新测评结果

  • [2024-08-16 16:35:23]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:3604kb
  • [2024-08-16 16:35:23]
  • 提交

answer

#include "bits/stdc++.h"
using namespace std;
using ll = long long;

const ll N = 1e6 + 1;
const ll M = 1e9 + 7;

ll n, k, avg;
string res;
string s;

string add(string str1, string str2) 
{ 
    // Remove leading zeros from both strings
    str1.erase(0, min(str1.find_first_not_of('0'), str1.size()-1));
    str2.erase(0, min(str2.find_first_not_of('0'), str2.size()-1));
 
    // If both strings become empty, return "0"
    if (str1.empty() && str2.empty()) 
        return "0";
 
    // Before proceeding further, make sure length 
    // of str2 is larger. 
    if (str1.length() > str2.length()) 
        swap(str1, str2); 
 
    // Take an empty string for storing result 
    string str = ""; 
 
    // Calculate length of both string 
    int n1 = str1.length(), n2 = str2.length(); 
    int diff = n2 - n1; 
 
    // Initially take carry zero 
    int carry = 0; 
 
    // Traverse from end of both strings 
    for (int i=n1-1; i>=0; i--) 
    { 
        // Do school mathematics, compute sum of 
        // current digits and carry 
        int sum = ((str1[i]-'0') + 
                (str2[i+diff]-'0') + 
                carry); 
        str.push_back(sum%10 + '0'); 
        carry = sum/10; 
    } 
 
    // Add remaining digits of str2[] 
    for (int i=n2-n1-1; i>=0; i--) 
    { 
        int sum = ((str2[i]-'0')+carry); 
        str.push_back(sum%10 + '0'); 
        carry = sum/10; 
    } 
 
    // Add remaining carry 
    if (carry) 
        str.push_back(carry+'0'); 
 
    // reverse resultant string 
    reverse(str.begin(), str.end()); 
 
    return str; 
} 

string cmin(string &a, string &b) {
    if (a.size() < b.size()) return a;
    if (a.size() > b.size()) return b;
    return (a < b) ? a : b;
}

void backtrack(ll pos, ll len, string sum, ll st) {
    if (pos > k) {
        if (len != n) return;
        res = cmin(res, sum);
        return;
    }
    for (int i = max(1ll, avg - 2); i <= avg + 2; i ++) {
        ll npos = pos + 1, nlen = len + i, nst = st + i;
        string nsum = sum;

        if (nlen > n) break;

        string t;
        for (int j = st; j <= st + i - 1; j ++) t += s[j];
        nsum = add(sum, t);
        backtrack(npos, nlen, nsum, nst);
    }
}

void doTest(ll testID) {
    cin >> n >> k; k ++;
    cin >> s; s = ' ' + s; 
    res = ""; for (int i = 1; i <= n; i ++) res += '9'; 
    avg = n / k;
    backtrack(1, 0, "0", 1); cout << res << '\n';
}

signed main() {
    ios_base::sync_with_stdio(0); cin.tie(0);

    int test = 1; 
    cin >> test;
    for (int _ = 1; _ <= test; _ ++) doTest(test);
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3604kb

input:

2
8 1
45455151
2 1
42

output:

9696
6

result:

ok 2 lines

Test #2:

score: -100
Time Limit Exceeded

input:

10
1301 6
56328399613959594774559774218276494124991536454496431869449134772679831477279356599352619469813771742358572734317965823527349354276551857226632977613336815474383422853946661428822284645652423563864641261338984158269966469425994769486371736593879954275146732544891889693921182364554588732946...

output:


result: