LEETCODE ALGORITHM:1155. Number of Dice Rolls With Target Sum

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1155. Number of Dice Rolls With Target Sum

You have n dice and each die has k faces numbered from 1 to k.

Given three integers n, k, and target, return the number of possible ways (out of the kn total ways) to roll the dice so the sum of the face-up numbers equals target. Since the answer may be too large, return it modulo 109 + 7.

Example 1:

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Input: n = 1, k = 6, target = 3
Output: 1
Explanation: You throw one die with 6 faces.
There is only one way to get a sum of 3.

Example 2:

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Input: n = 2, k = 6, target = 7
Output: 6
Explanation: You throw two dice, each with 6 faces.
There are 6 ways to get a sum of 7: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1.

Example 3:

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Input: n = 30, k = 30, target = 500
Output: 222616187
Explanation: The answer must be returned modulo 109 + 7.

Constraints:

  • 1 <= n, k <= 30
  • 1 <= target <= 1000

题解

\(dp[i][t]=d[i-1][t-1]+\dots+dp[i-1][t-k]\)

边界\(dp[1][j]=1,1<=j<=k\)

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class Solution {
public:
    int numRollsToTarget(int n, int k, int target) {
        // 状态定义
        // f[i][j]表示将i个筛子扔出和为j的方案数量，所以结果就是f[n][target]
        // max
        // 限制在于，每个筛子所能扔出的值范围是1-k
        const int mod = 1e9 + 7;
        int minN = min(k,target);
        int targetMax = n * k;
        vector<vector<int> > f(n+1,vector<int>(1001));

        // 定义边界条件
        // 表示用一个筛子 扔出min(k,target)的方案数量最多就一个
        for(int i = 1;i<=minN;i++){
            f[1][i]=1;
        }
        
        for(int i = 2;i<=n;i++){
            for(int j = i;j<=targetMax;j++){
                for(int l = 1;l<=j && l<=k;l++){ // 表示我这一轮扔出了l
                    f[i][j] = (f[i][j] + f[i - 1][j - l]) % mod;
                }
            }
        }
        return f[n][target];
    }
};


class Solution {
public:
    int numRollsToTarget(int n, int k, int target) {
        //vector<vector<int>> dp(n+1,vector<int>(1001));
        int dp[31][1001];
        memset(dp,0,sizeof(int)*31*1001);
        for(int i=1;i<=k;i++){
            dp[1][i]=1;
        }
        
        for(int i=2;i<=n;i++){
            for(int j=i;j<=target&&j<=i*k;j++){
                for(int l=1;l<=k&&j-l>=0;l++){
                    dp[i][j]=(dp[i][j]+dp[i-1][j-l])%1000000007;
                }
            }
        }
        return dp[n][target];
    }
};