ISBN号码
对于一个形式为“x-xxx-xxxxx-x”的ISBN，判断其是否合法。
题解
模拟，需要特别处理X。
#include <bits/stdc++.h>
using namespace std;
int main() {
char a[100];
scanf("%s", a);
int n = strlen(a);
int s = 0;
for (int i = 0, j = 0; i < n - 1; i++) {
if (a[i] != '-') {
j += 1;
s += (a[i] - '0') * j;
}
}
if (s % 11 == (a[n-1] - '0') || s % 11 == 10 && a[n-1] == 'X') {
puts("Right");
} else {
a[n-1] = s % 11 + '0';
if(s%11 == 10) {
a[n-1] = 'X';
}
puts(a);
}
}
排座椅
题解
贪心，分别统计每一行每一列加一条通道能隔开的数量，优先选择最多的K条横向的和L条纵向的。
#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
int m, n, k, l, d;
pair<int, int> row[N], col[N];
int main() {
cin >> m >> n >> k >> l >> d;
for (int i = 1; i <= max(n, m); i++) {
row[i].second = i;
col[i].second = i;
}
for (int i = 1; i <= d; i++) {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 == x2) {
col[min(y1, y2)].first += 1;
}
if (y1 == y2) {
row[min(x1, x2)].first += 1;
}
}
sort(row + 1, row + m + 1);
reverse(row + 1, row + m + 1);
sort(col + 1, col + n + 1);
reverse(col + 1, col + n + 1);
for (int i = 1; i <= k; i++) {
swap(row[i].first, row[i].second);
}
sort(row + 1, row + k + 1);
for (int i = 1; i <= k; i++) {
cout << row[i].first;
if (i != k) cout << ' ';
}
cout << endl;
for (int i = 1; i <= l; i++) {
swap(col[i].first, col[i].second);
}
sort(col + 1, col + l + 1);
for (int i = 1; i <= l; i++) {
cout << col[i].first;
if (i != l) cout << ' ';
}
}
传球游戏
有n个同学排成一个圈，从1号同学开始传球，每次可以向左传或向右传，问传m次回到1号同学有多少种方式。
题解
递推，用f[i][j]表示传了i次球，最后停留在j号同学的种数，转移方程为$$f[i][j] = f[i-1][j的左边] + f[i-1][j的右边]$$
#include <bits/stdc++.h>
using namespace std;
const int N = 35;
long long f[N][N];
int n, m;
int main() {
f[0][1] = 1;
cin >> n >> m;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
f[i][j] = (j == 1 ? f[i-1][n] : f[i-1][j-1]) + (j == n ? f[i-1][1] : f[i-1][j+1]);
}
}
cout << f[m][1] << endl;
}
立体图
题解
烂题。按从左往右，从前往后，从下往上的顺序画，之前存在的直接覆盖。
#include <bits/stdc++.h>
using namespace std;
const int N = 55;
int a[N][N];
int n, m;
char c[2005][2005];
char b[6][8] = {
"..+---+",
"./ /|",
"+---+ |",
"| | +",
"| |/.",
"+---+..",
};
int main() {
cin >> n >> m;
int mx = 0;
for (int i = 1; i <= n; i++) {
for (int j = 1 ;j <= m ; j++) {
scanf("%d", &a[i][j]);
mx = max(a[i][j], mx);
}
}
int px = 1e9, py = 0;
int qx = 0, qy = 0;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
for (int k = 1; k <= a[i][j]; k++) {
int x = 3 * (mx-k) + 2 * (i-1);
int y = 4 * (j-1) + 2 * (n-i);
px = min(px, x);
for(int p = 0; p < 6; p++) {
for(int q = 0; q < 7; q++) {
if (b[p][q] == '.') continue;
c[x + p][y + q] = b[p][q];
qx = max(qx, x + p);
qy = max(qy, y + q);
}
}
}
}
}
for (int i = px; i <= qx; i++) {
for (int j = 0; j <= qy; j++) {
if (c[i][j] == 0) cout << '.';else cout << c[i][j];
}
cout << endl;
}
}