【CF908G】New Year and Original Order

【CF908G】New Year and Original Order

题面

洛谷

题解

设(f[i][j][k][l])表示当前在第(i)位有(j)位大于等于(k)，当前有没有卡上界的方案数

则枚举新加的数(p)，有

[f[i+1][j+(pgeq k)][k][l|(p<a_i)]=sum f[i][j][k][l] ]

我们最后统计答案的时候枚举(k)

有

[ans=underbrace{111...11}_{j个1}*(f[i][j][k][0]+f[i][j][k][1]) ]

为什么要乘那么多(1)呢？(下面是张图片)

代码(压行有点丑)

#include <iostream> 
#include <cstdio> 
#include <cstdlib> 
#include <cstring> 
#include <cmath> 
#include <algorithm> 
using namespace std; 
#define rep(i, from, to) for(int i = (from); i <= (to); i++) 
const int Mod = 1e9 + 7; 
const int MAX_N = 1005; 
void pls(int &x, int y) { x += y; if (x >= Mod) x -= Mod; } 
char ch[MAX_N]; int a[MAX_N], N; 
int ans, f[MAX_N][MAX_N][10][2]; 
int main () { 
	scanf("%s", ch + 1); N = strlen(ch + 1); 
	rep(i, 1, N) a[i] = ch[i] - '0'; 
	rep(i, 0, 9) f[0][0][i][0] = 1; 
	rep(i, 0, N - 1) rep(j, 0, i) rep(k, 1, 9) rep(l, 0, 1) rep(p, 0, (l ? 9 : a[i + 1])) 
		pls(f[i + 1][j + (p >= k)][k][l | (p < a[i + 1])], f[i][j][k][l]); 
	rep(k, 1, 9) { 
		int res = 1; 
		rep(i, 1, N) pls(ans, 1ll * res * (f[N][i][k][0] + f[N][i][k][1]) % Mod), res = (10ll * res + 1) % Mod; 
	} 
	printf("%d
", ans); 
	return 0; 
}