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  • CodeForces 478D Red-Green Towers (DP)

    题意:给定 n 块红砖,m 块绿砖,问有多少种方式可以建造成最高的塔,每一层颜色必须一样。

    析:首先要确定最高是多少层h,大约应该是用 h * (h+1) <= (m+n) * 2,然后dp[i][j] 表示 前 i 层用 j 块红砖,dp[i][j] += dp[i-1][j-i],

    但是这个空间复杂度受不了,那么就变成滚动数组就好,dp[j] += dp[j-i],一个较简单的DP。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #include <sstream>
    #define debug() puts("++++");
    #define gcd(a, b) __gcd(a, b)
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const LL LNF = 1e17;
    const double inf = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 2e5 + 10;
    const int mod = 1000000007;
    const int dr[] = {-1, 0, 1, 0};
    const int dc[] = {0, 1, 0, -1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline bool is_in(int r, int c){
      return r >= 0 && r < n && c >= 0 && c < m;
    }
    
    int dp[maxn];
    
    int main(){
      scanf("%d %d", &n, &m);
      int h = (int)sqrt(n+m+m+n);
      while((h+1) * h <= n+n+m+m)  ++h;
      dp[0] = 1;
      for(int i = 1; i < h; ++i)
        for(int j = n; j >= i; --j)
          dp[j] = (dp[j] + dp[j-i]) % mod;
      int ans = 0;
      int all = h * (h-1) / 2;
      for(int i = n; i >= 0; --i){
        if(all - i > m)  break;
        ans = (ans + dp[i]) % mod;
      }
      printf("%d
    ", ans);
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7163419.html
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