n
个k
面的骰子,投掷出骰子的点数之和为target
的所有可能。
掷骰子等于目标和的方法数
动态规划,实际上相当于一个0-1背包。
令 d p [ i ] [ j ] dp[i][j] dp[i][j]为前 i i i个骰子和为j
的方案数
则
d p [ i ] [ j ] = ∑ t = 1 k d p [ i − 1 ] [ j − t ] , j ≤ t dp[i][j] = \sum_{t=1}^{k}dp[i - 1][j -t], j \le t dp[i][j]=t=1∑kdp[i−1][j−t],j≤t
class Solution {
public:
int numRollsToTarget(int n, int k, int target) {
// dp[i][j] = dp[i - 1][j - k] j - k >= i - 1
int dp[30 + 1][1000 + 1];
memset(dp, 0, sizeof(dp));
int MOD = 1e9 + 7;
dp[0][0] = 1;
for (int i = 1;i <= n; ++i)
for (int j = i + 1; j <= min(k * i, target); ++j) {
for (int inc = 1; inc <= k; ++inc ) {
if ( i - 1 <= j - inc )
dp[i][j] = (dp[i][j] + dp[i - 1][j - inc] )% MOD;
}
}
return dp[n][target];
}
};