Harbin University School of Software and Microelectronics Programming Contest (freshman season) - Synchronization game (review Bu Ti)

A - R does not simulate all the code and explain the focus of the last two questions

Lele small string S

https://ac.nowcoder.com/acm/contest/1877/S

C the number of records to traverse from front to back, showing there are several characters before the current character C. From the back through the records of the number N, N represents the number of characters after the character in the current

Finally, from front to back through the array, when the current character is H, there is numC * numN case, is the answer to every summed.

#include<bits/stdc++.h>
using namespace std;
const int MAXN=8000;
int num2[MAXN+11];
int main() {
    string s;
    cin>>s;
    for(int i=s.length()-1;i>=0;i--) {
        num2[i]=num2[i+1];
        if(s[i]=='N')
            num2[i]++;
    }
    long long cnt=0;
    for(int i=0;i<s.length();i++) {
        if(s[i]=='C') {
            for(int j=i+1;j<s.length();j++) {
                if(s[j]=='H')
                    cnt+=num2[j];
            }
        }
    }
    printf("%lld\n",cnt);
    return 0;
}

Lele small number of T Sec

https://ac.nowcoder.com/acm/contest/1877/T

Special record it, Portal

The first time I saw such a solution, number theory only GCD. %%%

#include<cstdio>  
#include<cstring>  
#include <algorithm>

using namespace std;

int ones[9999],tens[999], n, i, j, k, s, t, api, apj, aps, apt;
bool ck() {
    int p, r;
    if (i > 5)
        return 1;
    p = s;
    r = t;
    for (int q = 0; q < j; q++)
        p = p * 10 + s;
    for (int q = 0; q < i - j; q++)
        p = p * 10;
    for (int q = 1; q < i - j; q++)
        r = r * 10 + t;
    return p + r > n;
}
 
int main() {
    while (scanf("%d",&n), n) {
        printf("%d: ", n);
        if (n == 1) {
            puts("10");
            continue;
        }
        ones[0] = 1;
        tens[0] = 1;
        for (i = 1; i < 9999; i++)
            ones[i] = (ones[i - 1] * 10 + 1) % n;
        for (i = 1; i < 999; i++)
            tens[i] = tens[i - 1] * 10 % n;
        for (i = 1, aps = 0; i < 9999; i++) {
            k = 0;
            if ((n % 10 == 0 || n % 25 == 0) && i > 11)
                k = i - 11;
            for (j = k; j < i; j++)
                for (s = 1; s < 10; s++)
                    for (t = 0; t < (n % 10 ? 10 : 1); t++)
                        if (t != s && (((long long)ones[j]) * tens[i - j] * s + ones[i - j - 1] * t) % n == 0 && ck() &&
                            (!aps || s < aps || s == aps && j > apj && s < apt)) {
                            api = i;
                            apj = j;
                            aps = s;
                            apt = t;
                        }
            if (aps)
                break;
        }
        for (int x = 0; x < apj + 1; x++)
            printf("%d", aps);
        for (int x = 0; x < api - apj; x++)
            printf("%d", apt);
        printf("\n");
    }
    return 0;
}