1155. 掷骰子的N种方法

花了一个小时吧。。怪自己太菜了，还是把这道dp做了

 1 int numRollsToTarget(int d, int f, int target) {
 2     int mod = 7 + 1e9;
 3     if (d == 1 && f >= target)
 4         return 1;
 5     else if (d == 1 && f < target)
 6         return 0;
 7     vector<vector<int>> dp(d+1, vector<int>(target+1, 0));
 8     for (int i = 0; i <= target; i++)
 9         dp[1][i] = 0;
10 
11     for (int i = 1; i <= min(f,target); i++)
12         dp[1][i] = 1;
13     for (int i = 2; i <= d; i++)
14     {
15         for (int j = i; j <= min(i * f,target); j++)
16         {
17             for (int k = 1; k <= min(f, j-1); k++)
18             {
19                 dp[i][j] += dp[i - 1][j - k];
20                 dp[i][j] %= mod;
21             }
22         }
23     }
24     return dp[d][target];
25 }

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来源： https://www.cnblogs.com/zouma/p/11674001.html