Change: There are a number of unlimited coins, the value of 25 points, 10 points, 5 points and 1 points, please write code calculation n there are several representations.

Change of Change:

There are an unlimited number of coins, the value of which is 25, 10, 5 and 1, please write code to calculate n there are several representations.

Given an int n, there are several representations of how to return N. Ensure that n is less than or equal to 100000, in order to prevent overflow, please add the answer mod 1000000007.

Test examples

6

Returns: 2

Dynamic planning

Dp[i][sum] How many ways to make sum with I coins.

Set the face value of a v1,v2,v3,v4 coin

sum = v1*x1+ v2*x2 + v3*x3 + v4*k; Dp[i][sum]

{X1,x2,x3,k}

{0...X1}

{0...X2}

{0...X3}

{0...K}

k<=sum/v4;

Sum_0 = v1*x1+ v2*x2 + v3*x3 + v4* (k); Dp[i-1][sum-v4*k]

Sum_1 = v1*x1+ v2*x2 + v3*x3 + v4* (k-1); dp[i-1][sum-v4* (k-1)]

Sum_2 = v1*x1+ v2*x2 + v3*x3 + v4* (k-2); dp[i-1][sum-v4* (k-2)]

.

.

.

.

.

Sum_n = v1*x1+ v2*x2 + v3*x3 + v4* (k-n); dp[i-1][sum-v4* (K-n)]

.

.

.

Sum_k = v1*x1+ v2*x2 + v3*x3 +v4* (k-k); Dp[i-1][sum]

sum = Sum_0+sum_1+sum_2+sum_3+......+sum_n+sum_k;

Dp[i][sum]=dp[i-1][sum] + dp[i-1][sum-v4]+....dp[i-1][sum-v4*n] +dp[i-1][sum-v4*k]; k<=sum/v4;

Dp[i][sum] can be derived from the previous line

Code:

Two-dimensional arrays:

public int countways (int n) {

int coin[] = new int[]{1,5,10,25};

int dp[][] = new int[5][n+1];

for (int i = 0; I <= 4; ++i)

{

Dp[i][0] = 1;

}

for (int i=1;i<=4;++i) {

for (int j=1;j<=n;++j) {

for (int k=0;k<=j/coin[i-1];++k) {

Dp[i][j]+=dp[i-1][j-k*coin[i-1]];

}

}

}

return dp[4][n];

｝

Optimized for one-dimensional arrays:

Dp[i][sum]=dp[i-1][sum] + dp[i-1][sum-v4]+....dp[i-1][sum-v4*n] +dp[i-1][sum-v4*k]; k<=sum/v4;

Line: We just need the data from the previous row

Column: We only need the data before the current sum

One-dimensional arrays can be solved at a time by line

Suppose n=20

1 5 10 20

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

2 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5

3 1 1 1 1 1 2 2 2 2 2 4 4 4 4 4 4 4 4 4 4 9

4 1 1 1 1 1 2 2 2 2 2 4 4 4 4 4 4 4 4 4 4 9

public int countways (int n) {

int[] coin = new int[]{1,5,10,25};

Int[] dp = new INT[N+1];

Dp[0]=1;

for (int i = 0; i < coin.length; i++) {

for (int j = coin[i]; j <= N; j + +) {

DP[J] = (Dp[j]+dp[j-coin[i]])%1000000007;

}

}

return dp[n];

｝

Reference: http://www.cnblogs.com/python27/archive/2013/09/05/3303721.html title Source: http://www.nowcoder.com/practice/ c0503ca0a12d4256af33fce2712d7b24?tpid=8&tqid=11041&rp=3&ru=/ta/cracking-the-coding-interview& Qru=/ta/cracking-the-coding-interview/question-ranking