Submission #2540

#	Time	Username	Problem	Language	Result	Execution time	Memory
2540		tncks0121	생일수 II (GA4_birthday2)	C++98	100 / 100	85 ms	6168 KiB

birthday2

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <string>
#include <functional>
#include <vector>
#include <deque>
#include <utility>
#include <bitset>

using namespace std;
typedef long long ll;
const int INF = 987654321;
const ll LINF = (ll)1e15;

const int N = 200000;
const int N_ = N+10;
const ll MOD = 19980305;

struct mint {
    int v;
    mint(): v(0) { }
    mint(ll v): v((v+MOD*1000)%MOD) { }
    
    int operator*(){ return v; }
    
    mint operator+ (ll t) { return mint(v + t); }
    mint operator- (ll t) { return mint(v - t); }
    mint operator* (ll t) { return mint(v * t); }
    mint operator+ (mint t) { return mint(v + t.v); }
    mint operator- (mint t) { return mint(v - t.v); }
    mint operator* (mint t) { return mint((ll)v * t.v); }
};

mint operator+ (ll a, mint t) { return mint(a + t.v); }
mint operator- (ll a, mint t) { return mint(a - t.v); }
mint operator* (ll a, mint t) { return mint(a * t.v); }

mint E[N_];
mint A[N_], B[N_];
mint P3[N_], P10[N_];
mint T[N_];
char X[N_], Y[N_];

mint f (mint P, int n) {
    P = P * P10[n];
    mint ret = P3[n] - 1;
    ret = ret * P + B[n];
    ret = ret * P + T[n];
    return ret;
}

mint get (char *V) {
    int L = 0;
    mint ret, last, prefix;
    
    while(V[++L]); --L;
    
    for(int i = 1; i < L; i++) {
        ret = ret + last * (3 * E[i]) + T[i];
        last = 8 * E[i];
    }
    
    for(int i = 1; i <= L; i++) {
        if(V[i] >= '5') {
            mint low_prefix = prefix * 10 + 3;
            mint first = low_prefix * P10[L - i] + 3 * E[L - i];
            ret = ret + last * first + f(low_prefix, L - i);
            last = low_prefix * P10[L - i] + 8 * E[L - i];
        }
        if(V[i] >= '8') {
            mint low_prefix = prefix * 10 + 5;
            mint first = low_prefix * P10[L - i] + 3 * E[L - i];
            ret = ret + last * first + f(low_prefix, L - i);
            last = low_prefix * P10[L - i] + 8 * E[L - i];
        }
        prefix = prefix * 10 + (V[i] - '0');
    }
    
    ret = ret + last * prefix;
    return ret;
}

int main (){
    int i, j;
    
    P3[0] = P10[0] = 1;
    for(i = 1; i <= N; i++) {
        E[i] = E[i - 1] * 10 + 1;
        P3[i] = P3[i - 1] * 3;
        P10[i] = P10[i - 1] * 10;
        A[i] = A[i - 1] * 30 + 16 * P3[i - 1];
        B[i] = 2 * A[i] - 11 * E[i];
        T[i] = f(3, i - 1)
               + (3 * P10[i - 1] + 8 * E[i - 1]) * (5 * P10[i - 1] + 3 * E[i - 1]) + f(5, i - 1)
               + (5 * P10[i - 1] + 8 * E[i - 1]) * (8 * P10[i - 1] + 3 * E[i - 1]) + f(8, i - 1);
    }
    
    scanf("%s%s", X+1, Y+1);
    
    printf("%d\n", *(get(Y) - get(X)));
    return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 15 / 15
• Subtask 3: 20 / 20
• Subtask 4: 25 / 25
• Subtask 5: 30 / 30