5054. 拼接字符串
A,B,C一共有6中拼接,全部存储到一个数组里
然后针对每个查询,遍历数组中的每个拼接的字符串。
在上面这个过程中注意需要转换大小写,并且忽略非字母的符号
cpp
#include <stdio.h>
#include <cstring>
#include <iostream>
using namespace std;
#define ll long long
#define sf(x) scanf("%d", &x);
#define de(x) cout << x << " ";
#define Pu puts("");
const int N = 2e4 + 9;
string a, b, c;
string u[7];
string fun(string x) {
string res = "";
for (int i = 0; i < x.size(); i++) {
if (x[i] >= 'a' && x[i] <= 'z')
res += x[i];
else if (x[i] >= 'A' && x[i] <= 'Z')
res += (x[i] + 32);
}
return res;
}
int n;
int main() {
cin >> a >> b >> c;
a = fun(a);
b = fun(b);
c = fun(c);
for (int i = 0; i < 6; i++) {
u[i] = "";
}
u[0] = a + b + c;
u[1] = a + c + b;
u[2] = b + a + c;
u[3] = b + c + a;
u[4] = c + a + b;
u[5] = c + b + a;
cin >> n;
string t;
int flag;
while (n--) {
cin >> t;
t = fun(t);
flag = 0;
for (int i = 0; i < 6; i++) {
if (u[i] == t) {
flag = 1;
break;
}
}
cout << ((flag == 1) ? "ACC\n" : "WA\n");
}
return 0;
}
5055. 画矩形
这个题目参考的题解:
其实是从n-1条线中选2k条,从m-1条线中选2 k条
因为我们每次画一个举行,需要2条横边+2条竖边
这样对于我们选出来的(2*k)^2条边中,其实是从外往里画矩形
cpp
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define sf(x) scanf("%d", &x);
#define de(x) cout << x << " ";
#define Pu puts("");
const int N = 1e3 + 9, mod = 1e9 + 7;
int n, m;
int c[N][N];
int main() {
int k;
cin >> n >> m >> k;
for (int i = 0; i <= 1e3; i++) {
for (int j = 0; j <= i; j++) {
if (j == 0)
c[i][j] = 1;
else
c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
}
}
if (2 * k > n - 1 || 2 * k > m - 1)
cout << "0\n";
else
cout << ((ll)c[n - 1][2 * k] * (ll)c[m - 1][2 * k]) % mod;
// 注意这里需要进行强制转换,要不然1000,1000,250这个样例过不了
return 0;
}