杭电多校四

License Plate Recognition

题目太长就不放出来拉
题意
题目的意思是给一个车牌号，判断其中的每个汉字或字母或数字的开始和结束的区间
可以观察出其中每个字都会有一定的间隔，只需要特判汉字（因为汉字中有川和鄂）然后其他的通过双指针做就行了

#include<vector>
#include<string>
#include<iostream>
#include<memory.h>
using namespace std;
int t;
string s;
bool a[105];
int l, r, id = 0;
void solve() {
    id++;
    for (int i = 0; i < 100; i++)a[i] = 0;
    for (int i = 0; i < 30; i++) {
        cin >> s;
        for (int j = 0; j < 100; j++) {
            if (s[j] == '#')a[j] = 1;
        }
    }
    l = 0, r = 0;
    cout << "Case #" << id << ":" << endl;
    for (int i = 0; i < 20; i++) {
        if (l == 0 && a[i] == 1)l = i;
        if (a[i] == 1)r = i;
    }
    cout << l + 1 << " " << r + 1 << endl;
    l = 0, r = 0;
    for (int i = 21; i < 100; i++) {
        if (l == 0 && a[i] == 1)l = i;
        if (a[i] == 0 && a[i - 1] == 0 && l != 0) {
            r = i - 1;
            cout << l + 1 << " " << r << endl;
            l = 0, r = 0;
        }
    }
}
int main() {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cin >> t;
    while (t--)solve();
}

Kanade Loves Maze Designing

建图，然后跑一个dfs，最后根据题目所给公式快速幂计算结果

void dfs(int x,int f,int zz){
	w[stk][x]=zz;
	if(a[x].size()==1)return;
	for(auto i:a[x]){
		if(i==f)continue;
		if(st[c[i]]==0){
			st[c[i]]=1;
			dfs(i,x,zz+1);
			st[c[i]]=0;
		}
		else dfs(i,x,zz);
	}
}

快速幂

int qmi(int m, int k, int p)
{
    int res = 1 % p, t = m;
    while (k)
    {
        if (k&1) res = res * t % p;
        t = t * t % p;
        k >>= 1;
    }
    return res;
}