题意
给定 \(S \in ['>', '<']\)。表示排列 \(P\) 两点之间的大小关系。
求排列 \(P\) 的方案数。
Sol
排列方案,考虑 \(f_{i, j}\) 表示第 \(i\) 位的数在排列中排名为 \(j\) 的方案数。
- 当 \(S_i = '>'\),\(f_{i, j} = \sum_{k = 1} ^ {j - 1} f_{i - 1, k}\)。
- 当 \(S_i = '<'\),\(f_{i, j} = \sum_{k = j} ^ {i - 1} f_{i - 1, k}\)。
预处理是 \(trivial\) 的。
Code
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
using namespace std;
#ifdef ONLINE_JUDGE
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;
#endif
int read() {
int p = 0, flg = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') flg = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
p = p * 10 + c - '0';
c = getchar();
}
return p * flg;
}
string read_() {
string ans;
char c = getchar();
while (c != '<' && c != '>')
c = getchar();
while (c == '<' || c == '>') {
ans += c;
c = getchar();
}
return ans;
}
void write(int x) {
if (x < 0) {
x = -x;
putchar('-');
}
if (x > 9) {
write(x / 10);
}
putchar(x % 10 + '0');
}
const int N = 3005, mod = 1e9 + 7;
array <array <int, N>, N> f;
array <int, N> g;
void Mod(int &x) {
if (x >= mod) x -= mod;
if (x < 0) x += mod;
}
int main() {
int n = read();
string s = " " + read_();
f[1][1] = 1;
for (int i = 1; i <= n; i++) g[i] = 1;
for (int i = 2; i <= n; i++) {
for (int j = 1; j <= i; j++) {
if (s[i - 1] == '<') f[i][j] = g[j - 1];
else f[i][j] = g[i - 1] - g[j - 1];
Mod(f[i][j]);
}
for (int j = 1; j <= i; j++)
g[j] = g[j - 1] + f[i][j], Mod(g[j]);
}
write(g[n]), puts("");
return 0;
}
标签:DPT,int,ans,include,Permutation,mod,getchar
From: https://www.cnblogs.com/cxqghzj/p/17850799.html