srm 555

欢迎点此阅读QvQ

255

Description

问能将一个0/10/1串最少分成几个串使得每个串都是55的整数次幂，且没有前导零

Solution

dp即可，dp[i]dp[i]表示前ii个字符分割的结果，枚举一个分割点jj，checkcheck即可

Code

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define F first
#define S second
typedef long long LL;
typedef pair<int, int> pii;
const int N = 55;
int dp
;
struct CuttingBitString {
bool ok(LL x) {
while (x && x % 5 == 0) x /= 5;
return x == 1;
}
int getmin(string S) {
string s = "0" + S + "1";
int n = s.size();
for (int i = 1; i <= n; ++i)    dp[i] = 10000;
for (int i = 1; i < n; ++i) {
if (s[i] == '1') {
for (int j = 0; j < i - 1; ++j) {
if (dp[j] == 10000 || s[j + 1] == '0')  continue;
LL now = 0;
for (int k = j + 1; k <= i - 1; ++k)    now = now * 2 + (s[k] == '1' ? 1 : 0);
if (ok(now))    dp[i - 1] = min(dp[i - 1], dp[j] + 1);
}
}
}
return dp[n - 2] == 10000 ? -1 : dp[n - 2];
}
};

555

Description

有一个H∗WH*W的全00矩阵，给定对行列操作数，每次操作将行列0/10/1翻转，使得最后矩阵11的个数为SS,求方案数

两个方案不同当且仅当有一行或列操作数不同

Solution

很容易想到操作两次相当于没操作，枚举行列分别操作11次的个数，剩下的相当于nn个物品(n对操作两次)放到mm个不同盒子的方案数，即(n+m−1n)\binom{n+m-1}{n}，统计即可

Code

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define F first
#define S second
typedef long long LL;
typedef pair<int, int> pii;
const int N = 1600, M = 555555555;
int c[N << 1][N << 1];
struct XorBoard {
int count(int H, int W, int Rcount, int Ccount, int S) {
for (int i = 0; i <= 1555 * 2; ++i) {
c[i][0] = 1;
for (int j = 1; j <= i; ++j)    c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % M;
}
int ans = 0;
for (int i = 0; i <= Rcount && i <= H; ++i)
for (int j = 0; j <= Ccount && j <= W; ++j) {
if (i * W + j * H - 2 * i * j != S) continue;
if ((Rcount - i) & 1 || (Ccount - j) & 1)   continue;
int rl = (Rcount - i) / 2, cl = (Ccount - j) / 2;
(ans += (LL)c[rl + H - 1][rl] * c[H][i] % M * c[cl + W - 1][cl] % M * c[W][j] % M) %= M;
}
return ans;
}
};