感覺自己太頹廢了,於是想寫點東西來激勵自己
於是,記錄下NOI 2020前的每天
前言:由於GD的yyc,ntf十分好人,送了我個E1,GD的各位老哥好心地送了我個rk10
但我深知自己的實力遠遠不及GD省隊。
不過進了,那就好好拼上一把吧
##2020-7-2
看了看agc 046D
感覺xyx大佬的res數組可以排除一些dp檢測不到的地方?
改了一下,把一些地方編了一下,然後過了(
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN = 305;
const int P = 998244353;
typedef long long ll;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); }
template <typename T> void read(T &x) {
x = 0; int f = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
x *= f;
}
char s[MAXN]; int n;
int dp[MAXN][MAXN][MAXN];
int vis[MAXN][MAXN][MAXN];
bool res[MAXN][MAXN][MAXN];
void update(int &x, int y) {
x += y;
if (x >= P) x -= P;
}
bool check(int x, int y, int z) {
return res[x][y][z];
}
int main() {
scanf("%s", s + 1), n = strlen(s + 1), dp[n][0][0] = 1;
for (int i = n; i >= 1; i--)
for (int j = 0; j + (n - i) <= n - 1; j++)
for (int k = 0; j + k + (n - i) <= n - 1; k++)
if (s[i] == '0') {
update(dp[i - 1][j][k], dp[i][j][k]);
update(dp[i][j][k + 1], dp[i][j][k]);
} else {
update(dp[i - 1][j][k], dp[i][j][k]);
update(dp[i][j + 1][k], dp[i][j][k]);
}
memset(vis, -1, sizeof(vis));
vis[0][0][0] = 0;
for (int i = 0; i <= n; i++)
for (int j = 0; j <= i / 2; j++)
for (int k = 0; j + k <= i / 2; k++) {
if (vis[i][j][k] == -1) continue;
int tmp = vis[i][j][k];
if (i + 1 <= n && tmp != 0) {
if (s[i + 1] == '0') chkmax(vis[i + 1][j + 1][k], tmp - 1);
if (s[i + 1] == '1') chkmax(vis[i + 1][j][k + 1], tmp - 1);
chkmax(vis[i+1][j][k],tmp);
}
if (i + 2 <= n) {
if (s[i + 1] == '0' || s[i + 2] == '0') chkmax(vis[i + 2][j + 1][k], tmp);
if (s[i + 1] == '1' || s[i + 2] == '1') chkmax(vis[i + 2][j][k + 1], tmp);
chkmax(vis[i + 2][j][k], tmp + 1);
chkmax(vis[i+1][j][k],tmp);
}
}
for (int i = 0; i <= n; i++)
for (int j = n; j >= 0; j--)
for (int k = n; k >= 0; k--) {
res[i][j][k] = vis[i][j][k] != -1;
/* if (i != 0) res[i][j][k] |= res[i - 1][j][k];
res[i][j][k] |= res[i][j + 1][k];
res[i][j][k] |= res[i][j][k + 1];*/
}
int ans = 0;
for (int i = 0; i <= n; i++)
for (int j = 0; j + (n - i) <= n; j++)
for (int k = 0; j + k + (n - i) <= n; k++)
if (dp[i][j][k] && (n - i) + j + k != 0) {
if (check(i, j, k)) {
//cout<<"i j k "<<i<<" "<<j<<" "<<k<<endl;
update(ans, dp[i][j][k]);
}
}
cout << ans << endl;
return 0;
}