又熬夜刷了cf，今天比正常多一题，比赛还没完但我知道F过不了了，一个半小时贡献给F还是没过……应该也没人Hack，写写解题报告吧= =！
解题报告如下：
A题：选择排序直接搞，因为不要求最优交换次数，代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <memory.h>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <string>
#include <cstring>
using namespace std;
#define Clear(f, nr) memset(f, nr, sizeof(f))
const int SIZE = 3001;
const int MSIZE = 10000;
const int INF = 1 << 30;
typedef long long ll;
pair<int, int> p[SIZE];
int main() {
int n;
int a[SIZE];
while(cin >> n) {
for(int i = 0; i < n; i ++)
cin >> a[i];
int k = 0;
for(int i = 0; i < n - 1; i ++) {
int Mi = a[i];
int mark = i;
for(int j = i + 1; j < n; j ++) {
if(a[j] < Mi) {
Mi = a[j];
mark = j;
}
}
if(mark != i) {
swap(a[i], a[mark]);
p[k ++] = make_pair(i, mark);
}
}
cout << k << endl;
for(int i = 0; i < k; i ++)
cout << p[i].first << " " << p[i].second << endl;
}
}
B题：贪心思想，排序后从小到大匹配即可：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <memory.h>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <string>
#include <cstring>
using namespace std;
#define Clear(f, nr) memset(f, nr, sizeof(f))
const int SIZE = 500;
const int MSIZE = 10000;
const int INF = 1 << 30;
typedef long long ll;
int main() {
int n, m;
int a[SIZE], b[SIZE];
while(cin >> n) {
for(int i = 0; i < n; i ++)
cin >> a[i];
cin >> m;
for(int i = 0; i < m; i ++)
cin >> b[i];
sort(a, a + n);
sort(b, b + m);
int sum = 0;
bool vis[SIZE];
Clear(vis, 0);
for(int i = 0; i < n; i ++) {
for(int j = 0; j < m; j ++) {
if(!vis[j] && abs(a[i] - b[j]) <= 1) {
vis[j] = 1;
sum ++;
break;
}
}
}
cout << sum << endl;
}
}
C题：贪心思想，如果是最小值，除去第一位尽可能放0，第一位尽可能放1。同理，最大值尽可能放9
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <memory.h>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <string>
#include <cstring>
using namespace std;
#define Clear(f, nr) memset(f, nr, sizeof(f))
const int SIZE = 500;
const int MSIZE = 10000;
const int INF = 1 << 30;
typedef long long ll;
string gMi(int m, int s) {
string ans;
int tmp = s - 9 * (m - 1);
if(tmp <= 0) {
ans += '1';
s -= 1;
}
else {
ans += tmp + '0';
s -= tmp;
}
for(int i = 1; i < m; i ++) {
int tmp = s - 9 * (m - i - 1);
if(tmp <= 0) {
ans += '0';
}
else {
ans += tmp + '0';
s -= tmp;
}
}
return ans;
}
string gMx(int m, int s) {
string ans;
for(int i = 0; i < m; i ++) {
if(s >= 9) {
ans += '9';
s -= 9;
}
else {
ans += s + '0';
s = 0;
}
}
return ans;
}
int main() {
int m, s;
while(cin >> m >> s) {
if(s > 9 * m || (m != 1 && s == 0)) {
puts("-1 -1");
continue;
}
if(m == 1 && s == 0) {
puts("0 0");
continue;
}
string Mi = gMi(m, s);
string Mx = gMx(m, s);
cout << Mi << " " << Mx << endl;
}
}
D题：读题花了好久，题目意思就是找菱形（4个点构成），思路就是dfs2层，对于最后一层如果x点的入度为y，则x点构成的菱形个数为y * (y - 1) / 2
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <memory.h>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <string>
#include <cstring>
using namespace std;
#define Clear(f, nr) memset(f, nr, sizeof(f))
const int SIZE = 3030;
const int MSIZE = 10000;
const int INF = 1 << 30;
typedef long long ll;
vector<int> path[SIZE];
ll ans;
int in[SIZE];
void dfs(int x, int root, int flag) {
if(flag == 0) {
if(x != root)
in[x] ++ ;
return ;
}
for(int i = 0; i < path[x].size(); i ++) {
int son = path[x][i];
dfs(son, root, flag - 1);
}
return ;
}
int main() {
int n, m;
int x, y;
while(cin >> n >> m) {
for(int i = 0; i < n; i ++)
path[i].clear();
for(int i = 0; i < m; i ++) {
cin >> x >> y;
x --,y --;
path[x].push_back(y);
}
ans = 0;
for(int i = 0; i < n; i ++) {
Clear(in, 0);
dfs(i, i, 2);
for(int j = 0; j < n; j ++) {
//printf("i:%d, j:%d -> %d\n", i, j, in[j]);
if(in[j] >= 2)
ans += in[j] * (in[j] - 1) / 2;
}
}
cout << ans << endl;
}
}
E题：没读
F题：看了别人的报告。思路就是：记录每列1的个数，令nr0代表有多少列含有0个1，nr1代表有多少列含1个1。因为和位置无关，递推方程式：
for i,j: [i][j] += [i-2][j+2] * C(i,2) [i][j] += [i-1][j-1+1] * C(i,1)*C(j,1) [i][j] += [i][j-2] * C(j,2)
最后输出f[0][0]，代表全部填充完毕
由于递推方程式不太好在循环中实现，就改为记忆化搜索，代码如下：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <memory.h>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <string>
#include <cstring>
using namespace std;
#define Clear(f, nr) memset(f, nr, sizeof(f))
const int SIZE = 502;
const int MSIZE = 10000;
const int INF = 1 << 30;
typedef long long ll;
ll rec[SIZE][SIZE];
int n, m, mod, x;
ll dp(int nr0, int nr1) {
if(rec[nr0][nr1] != -1)
return rec[nr0][nr1];
if(nr0 == nr1 && nr0 == 0)
return rec[0][0] = 1;
ll ans = 0;
if(nr0 >= 2) {
ll tmp = dp(nr0 - 2, nr1 + 2) % mod;
tmp = tmp * ((nr0 * (nr0 - 1)) / 2 % mod) % mod;
ans = (ans + tmp) % mod;
}
if(nr1 >= 2) {
ll tmp = dp(nr0, nr1 - 2) % mod;
tmp = tmp * ((nr1 * (nr1 - 1)) / 2 % mod) % mod;
ans = (ans + tmp) % mod;
}
if(nr1 >= 1 && nr0 >= 1) {
ll tmp = dp(nr0 - 1, nr1);
tmp = tmp * ((nr0 * nr1) % mod) % mod;
ans = (ans + tmp) % mod;
}
return rec[nr0][nr1] = ans;
}
int main() {
int col[SIZE];
while(cin >> n >> m >> mod) {
Clear(col, 0);
Clear(rec, -1);
for(int i = 0; i < m; i ++)
for(int j = 0; j < n; j ++) {
scanf("%1d", &x);
col[j] += x;
}
int nr0 = 0, nr1 = 0;
for(int j = 0; j < n; j ++) {
if(col[j] == 0) nr0 ++;
else if(col[j] == 1) nr1 ++;
}
cout << dp(nr0, nr1) << endl;
}
}