[CF118D]Caesar's Legions 题解

题意简述

一个01序列由\(n_1\)个0和\(n_2\)个1组成，求最长连续0串长度不超过\(k_1\)，最长连续1串长度不超过\(k_2\)的序列的方案总数

题解

状态

方案总数

变量

已经取了i个0，j个1，当前末尾连续串长度为k，末尾为l。

转移

\[f[i][j][k][l] = \left\{ \begin{matrix} \sum_{x=1}^{min(j,k_2)} f[i-[l=0]][j-[l=1]][x][l\ xor\ 1] && k = 1\\ f[i-[l=0]][j-[l=1]][k-1][l] && k > 1\\ \end{matrix} \right. \]

 注:\([i=1]\)意为在\(i=1\)时值为\(1\)，否则值为\(0\)。

代码

#include <cstdio>
#include <algorithm>

using namespace std;

const long long MOD = 100000000;

namespace fast_IO{
    const int IN_LEN = 10000000, OUT_LEN = 10000000;
    char ibuf[IN_LEN], obuf[OUT_LEN], *ih = ibuf + IN_LEN, *oh = obuf, *lastin = ibuf + IN_LEN, *lastout = obuf + OUT_LEN - 1;
    inline char getchar_(){return (ih == lastin) && (lastin = (ih = ibuf) + fread(ibuf, 1, IN_LEN, stdin), ih == lastin) ? EOF : *ih++;}
    inline void putchar_(const char x){if(oh == lastout) fwrite(obuf, 1, oh - obuf, stdout), oh = obuf; *oh ++= x;}
    inline void flush(){fwrite(obuf, 1, oh - obuf, stdout);}
    int read(){
        int x = 0; int zf = 1; char ch = ' ';
        while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar_();
        if (ch == '-') zf = -1, ch = getchar_();
        while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar_(); return x * zf;
    }
    void write(int x){
        if (x < 0) putchar_('-'), x = -x;
        if (x > 9) write(x / 10);
        putchar_(x % 10 + '0');
    }
}

using namespace fast_IO;

long long f[105][105][11][2];

int main(){
	int n1 = read(), n2 = read(), k1 = read(), k2 = read();
	for (int i = 1; i <= k1; ++i) f[i][0][i][0] = 1;
	for (int i = 1; i <= k2; ++i) f[0][i][i][1] = 1;
	for (int i = 1; i <= n1; ++i)
		for (int j = 1; j <= n2; ++j){
			for (int k = 1; k <= min(j, k2); ++k)
				(f[i][j][1][0] += f[i - 1][j][k][1]) %= MOD;
			for (int k = 1; k <= min(i, k1); ++k)
				(f[i][j][1][1] += f[i][j - 1][k][0]) %= MOD;
			for (int k = 2; k <= min(i, k1); ++k)
				(f[i][j][k][0] += f[i - 1][j][k - 1][0]) %= MOD;
			for (int k = 2; k <= min(j, k2); ++k)
				(f[i][j][k][1] += f[i][j - 1][k - 1][1]) %= MOD;
		}
	long long ans = 0;
	for (int i = 1; i <= 10; ++i)
		(ans += f[n1][n2][i][0] + f[n1][n2][i][1]) %= MOD;
	printf("%lld", ans);
	return 0;
}