喜欢将平衡比作钟摆的这个形容
links
https://github.com/ssoccean/PPT_Pilot
reinforcemennt learning需要考虑的核心问题之一
掌握好explore和exploit的接受平衡非常重要
idea-实践 循环
核心就两件事
用工程进度文件而不是上下文压缩长时工作更好
规划、执行、评估分开更好能解决模型自信问题
lc1220
class Solution {
public:
vector<long long> state;
// a e i o u
// 0 1 2 3 4
// 01 10 12 20 21 23 24 32 34 40
long long mod = 1e9 + 7;
vector<long long> run(vector<long long>& s) {
vector<long long> s_(5);
s_[0] = (s[1] + s[2] + s[4]) % mod;
s_[1] = (s[0] + s[2]) % mod;
s_[2] = (s[1] + s[3]) % mod;
s_[3] = s[2] % mod;
s_[4] = (s[2] + s[3]) % mod;
return s_;
}
int countVowelPermutation(int n) {
state = vector<long long>(5, 1);
for (int i = 1; i < n; i++) {
++state = run(state);++
}
long long sum = 0;
for (int i = 0; i < 5; i++) {
++sum += state[i];++
sum %= mod;
}
return sum;
}
};
lc1269
归来任是第一反应写最朴素的dp 释怀的笑了
const int MO = 1e9 + 7;
class Solution {
public:
int numWays(int steps, int arrLen) {
vector<vector<long long>> f(steps + 1, vector<long long>(steps + 1));
long long maxLen = min(steps - 1, arrLen - 1);
f[0][0] = 1;
for (int i = 1; i <= steps; ++i) {
for (int j = 0; j <= maxLen; ++j) {
f[i][j] = f[i - 1][j];
if (j - 1 >= 0)
f[i][j] = (f[i][j] + f[i - 1][j - 1]) % MO;
if (j + 1 <= maxLen)
f[i][j] = (f[i][j] + f[i - 1][j + 1]) % MO;
}
}
return f[steps][0] % MO;
}
};