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数位dp

Author: tmp-coder

leetcode/600.不含连续-1-的非负整数.cpp

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+/*
+ * @lc app=leetcode.cn id=600 lang=cpp
+ *
+ * [600] 不含连续1的非负整数
+ */
+#include<bits/stdc++.h>
+using namespace std;
+// @lc code=start
+class Solution {
+public:
+    int solve(int len, int last, int over, vector<vector<vector<int>>>&dp, bitset<32> & dim){
+        if(len == 32)
+            return 1;
+        if(dp[len][last][over] != -1)
+            return dp[len][last][over];
+        
+        auto & ans = dp[len][last][over];
+        ans = 0;
+        int k = dim[31-len];
+        int maxx = over ? k : 1 ;
+        if(last==1)
+            maxx = 0;
+        for(int i=0; i<=maxx; ++i){
+            ans += solve(len + 1, i, over && (i==k), dp, dim);
+        }
+        return ans;
+    }
+    int findIntegers(int num) {
+        vector<vector<vector<int> > > dp(32,vector<vector<int>>(2,vector<int>(2,-1)));
+        bitset<32> dim(num);
+        
+        // for(int i=0 ; i<32; ++i)
+        //     cout << dim[i] << "\n";
+        // cout << dim << "\n";
+        return solve(0, 0,1, dp, dim);
+    }
+};
+// @lc code=end
+