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; }