PAT顶级 1025 Keep at Most 100 Characters (35分)

题目链接：

1025 Keep at Most 100 Characters (35分)

思路：

用dp[i][j]记录字符串从第一个字符到第i个字符形成的子串、保存j个字符所形成的sub-sequence个数；
则我们可以得到递推式$dp[i][j]=dp[i-1][j]+dp[i-1][j-1]$
但是这种思想肯定会造成重复的！
记此时char c = s[i]，则我们需要找到上一次c出现的位置pos，减去$dp[pos-1][j-1]$即可~

代码：

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;

const ll mod = 1000000007;
const int maxn = 1005;
const int maxm = 105;
int rcd[30];
ll dp[maxn][maxm];
string s;
void solve(){
	s = "0" + s;
	int n = s.length() - 1;
	for(int i = 1; i <= n; i++){
		int pos = rcd[s[i] - 'a'];
		dp[i][1] = dp[i - 1][1] + !pos;
		for(int j = 2; j < maxm; j++){
			dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1] - (pos ? dp[pos - 1][j - 1] : 0) + mod) % mod;
		}
		rcd[s[i] - 'a'] = i;
	}
	ll ans = 0;
	for(int i = 1; i <= 100; i++) ans = (ans + dp[n][i]) % mod;
	cout << ans;
}
int main(){
#ifdef LOCAL
	freopen("Sakura.txt", "r", stdin);
#endif
	cin >> s;
	solve();
	return 0;
}