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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
#580269#8932. BingoForever_Young#WA 67ms6000kbC++176.2kb2024-09-21 20:49:322024-09-21 20:49:33

Judging History

你现在查看的是最新测评结果

  • [2024-09-21 20:49:33]
  • 评测
  • 测评结果:WA
  • 用时:67ms
  • 内存:6000kb
  • [2024-09-21 20:49:32]
  • 提交

answer

#include<vector>
#include<set>
#include<map>
#include<queue>
#include<string>
#include<algorithm>
#include<iostream>
#include<bitset>
#include<functional>
#include<chrono>
#include<numeric>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cassert>
#include<cmath>
#include<iomanip>
#include<random>
#include<ctime>
#include<complex>
#include<type_traits>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef long long LL;
typedef double D;
#define all(v) (v).begin(), (v).end()
mt19937 gene(chrono::system_clock::now().time_since_epoch().count());
typedef complex<double> Complex;
typedef pair<int, int> pii;
#define fi first
#define se second
#define ins insert
#define pb push_back
inline char GET_CHAR(){
    const int maxn = 131072;
    static char buf[maxn],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,maxn,stdin),p1==p2)?EOF:*p1++;
}
inline int getInt() {
    int res(0);
    char c = getchar();
    while(c < '0') c = getchar();
    while(c >= '0') {
        res = res * 10 + (c - '0');
        c = getchar();
    }
    return res;
}
inline LL fastpo(LL x, LL n, LL mod) {
    LL res(1);
    while(n) {
        if(n & 1) {
            res = res * (LL)x % mod;
        }
        x = x * (LL) x % mod;
        n /= 2;
    }
    return res;
}
template<LL mod> struct Num {
    LL a;
    Num operator + (const Num & b) { return Num{(a + b.a) % mod}; }
    Num operator - (const Num & b) { return Num{(a - b.a + mod) % mod}; }
    Num operator * (const Num & b) { return Num{a * b.a % mod}; }
    Num operator / (const Num & b) { return Num{a * fastpo(b.a, mod - 2, mod) % mod}; }
    void operator += (const Num & b) {if((a += b.a) >= mod) a -= mod;}
    void operator -= (const Num & b) {if((a -= b.a) < 0) a += mod;}
    void operator *= (const Num & b) { a = a * b.a % mod; }
    void operator /= (const Num & b) { a = a * fastpo(b.a, mod - 2, mod) % mod; }
    void operator = (const Num & b) { a = b.a; }
    void operator = (const LL & b) { a = b; }
};
template<LL mod> ostream & operator << (ostream & os, const Num<mod> & a) {
    os << a.a;
    return os;
}
LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a; }
inline string itoa(LL x, int width = 0) {
    string res;
    if(x == 0) res.push_back('0');
    while(x) {
        res.push_back('0' + x % 10);
        x /= 10;
    }
    while((int)res.size() < width) res.push_back('0');
    reverse(res.begin(), res.end());
    return res;
}
const int _B = 131072;
char buf[_B];
int _bl = 0;
inline void flush() {
    fwrite(buf, 1, _bl, stdout);
    _bl = 0;
}
__inline void _putchar(char c) {
    if(_bl == _B) flush();
    buf[_bl++] = c;
}
inline void print(LL x, char c) {
    static char tmp[20];
    int l = 0;
    if(!x) tmp[l++] = '0';
    else {
        while(x) {
            tmp[l++] = x % 10 + '0';
            x /= 10;
        }
    }
    for(int i = l - 1; i >= 0; i--) _putchar(tmp[i]);
    _putchar(c);
}
typedef double C;
struct P {
    C x, y;
    void scan() {
        double _x, _y;
        scanf("%lf%lf", &_x, &_y);
        x = _x; y = _y;
    }
    void print() {
        cout << '(' << x << ' ' << y << ')' << endl;
    }
    P operator + (const P & b) const { return P{x + b.x, y + b.y}; }
    P operator - (const P & b) const { return P{x - b.x, y - b.y}; }
    C operator * (const P & b) const { return x * b.y - y * b.x; }
    C operator % (const P & b) const { return x * b.x + y * b.y; }
};
P operator * (const C & x, const P & b) { return P{x * b.x, x * b.y}; }
const int N = 1000133;
const int LOG = 20;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int rela[N];
int getr(int x) {
    int p = x;
    while(rela[p] != p) p = rela[p];
    int p1 = p; p = x;
    while(rela[p] != p) {
        int p2 = rela[p];
        rela[p] = p1;
        p = p2;
    }
    return p1;
}

char st[N];
int n[N];
LL toi(int * st, int * ed) {
    // printf("%p %p\n", st, ed);
    LL res = 0;
    for (int * p = ed - 1; ; p--) {
        res = res * 10 + (*p);
        if (p == st) {
            break;
        }
    }
    return res;
}
void solve() {
    int m;
    scanf("%s", st);
    scanf("%d", &m);
    // printf("!!\n");
    int len = strlen(st);
    LL rem = 0;
    
    // printf("!!\n");
    for (int i = 0; i < len; i++) {
        n[i] = st[i] - '0';
    }
    reverse(n, n + len);
    for (int i = len - 1; i >= 0; i--) {
        rem = (rem * 10 + n[i]) % m;
    }
    
    // printf("!!\n");
    int mlen = 0;
    int x = m;
    LL d = 1;
    LL pw = 1;
    
    // printf("!!\n");
    while (x) {
        // xm[mlen++] = x % 10;
        mlen++;
        d *= 10;
        x /= 10;
    }
    for (int i = 0; i <= mlen + 10; i++) {
        n[len++] = 0;
    }
    
    LL mn = inf;
    mn = rem == 0 ? m : m - rem;
    LL sat = 1;
    // printf("??\n");
    for (int i = 0; i + mlen <= len; i++) {
        // printf("i = %d\n", i);
        LL frm = toi(n + i, n + i + mlen);
        LL to = m;
        // printf("%d %lld %lld\n", i, frm, to);
        // printf("i = %d\n", i);
        if (frm == to) {
            mn = 0;
            break;
        }
        LL delta;
        if (frm < to) {
            delta = (to - frm - 1) * pw + sat;
        } else {
            delta = (to + d - frm - 1) * pw + sat;
        }
        if (delta < mn) {
            mn = delta;
        }
        sat += (9 - n[i]) * pw;
        if (sat > inf) {
            sat = inf;
        }
        pw = pw * 10;
        if (pw > inf) {
            pw = inf;
        }
    }
    for (int i = 0; i < len; i++) {
        mn += n[i];
        n[i] = mn % 10;
        mn /= 10;
    }
    while (n[len - 1] == 0) {
        len--;
    }
    for (int i = len - 1; i >= 0; i--) {
        printf("%d", n[i]);
    }
    printf("\n");
}
int main() {
    int t;
    scanf("%d", &t);
    for (int qq = 1; qq <= t; qq++) {
        solve();
    }
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 5864kb

input:

6
7 3
12 3
9 10
249 51
1369 37
2 1

output:

9
13
10
251
1370
3

result:

ok 6 lines

Test #2:

score: -100
Wrong Answer
time: 67ms
memory: 6000kb

input:

100000
3196282243 28
7614814237 33
2814581084 97
1075124401 58
7822266214 100
1767317768 31
7189709841 75
9061337538 69
6552679231 38
9946082148 18
5497675062 54
7787300351 65
4310767261 68
4811341953 100
3265496130 31
8294404054 62
2845521744 90
1114254672 26
6442013672 13
3744046866 40
3289624367 ...

output:

3196282243
7614814251
2814581097
1075124424
7822266300
1767317768
7189709850
9061337569
6552679238
9946082160
5497675062
7787300365
4310767268
4811342000
3265496131
8294404062
2845521790
1114254674
6442013672
3744046866
3289624375
6477935360
1292587551
5504674689
2898829180
7882736025
2846033387
923...

result:

wrong answer 1st lines differ - expected: '3196282244', found: '3196282243'