Va 674-Coin Change

There are five types of coins: 1, 5, 10, 25, 50. Now let's get a number of n. What is the total number of solutions to combine these five coins and n?

Solution: Dynamic Planning
1: If a coin consists of only one cent, the total number of coins is 1.
2: If a coin is composed of 1 and 5, the total number of solutions is 1 + s1 (s1 is a solution consisting of 1 and 5)
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6: If a coin consists of, 25, and 50, the total number of coins is 1 + s1 + s2 + s3 + s4 + s5.
The maximum number of data n given in the question is 7489, so we first calculate the total number of solutions to split all values (that is, to create a table ), then input a me file and output it directly.
Idea: we use the type [5] array to store 5 types of coins, and set dp [I] [j] to indicate the first I + 1 type (type [0]... type [I]) is the total number of solutions whose value is j. Then I can enumerate every value j in two cycles, find all the data, and finally find the output.

Code:
[Cpp]
// (No optimization)
# Include <algorithm>
# Include <iostream>
# Include <cstring>
# Include <string>
# Include <vector>
# Include <cstdio>
# Include <stack>
# Include <queue>
# Include <cmath>
Using namespace std;
# Define maxn8000
 
Int type [5] = {1, 5, 10, 25, 50 };
Int dp [5] [MAXN]; // it can be of the int type. If MAXN is tens of thousands, use unsigned long
Int n, ans;
 
// Process all data in a table
Void solve (){
For (int I = 0; I <5; I ++ ){
For (int j = 0; j <MAXN; j ++ ){
If (I = 0 | j = 0) dp [I] [j] = 1; // If I is 0, it indicates that only 1 is composed of dp [I] [j] 1, if j is 0, it indicates that there is only one composition scheme with a value of 0 (if it is regarded as 0, it will WA)
Else dp [I] [j] = 0; // all others are initialized to 0
}
}
For (int I = 1; I <5; I ++) {// The I-1 appears, then enumerate from 1
For (int j = 1; j <MAXN; j ++ ){
For (int k = 0; k * type [I] <= j; k ++) // k types [I]. Note that the value must start from 0.
Dp [I] [j] + = dp [I-1] [j-k * type [I]; // Add the number of preceding solutions
}
}
}
 
Int main (){
// Freopen ("input.txt", "r", stdin );
Solve ();
While (scanf ("% d", & n )! = EOF ){
For (int I = 4; I> = 0; I --){
If (dp [I] [n]) {ans = dp [I] [n]; break;} // search from the end. if either of them is not 0, the answer is returned.
}
Printf ("% d \ n", ans );
}
Return 0;
}
 
 
 
 
// (Optimize two-dimensional data into one-dimensional data) Optimized Code
# Include <algorithm>
# Include <iostream>
# Include <cstring>
# Include <string>
# Include <vector>
# Include <cstdio>
# Include <stack>
# Include <queue>
# Include <cmath>
Using namespace std;
# Define maxn8000
 
Int type [5] = {1, 5, 10, 25, 50 };
Int dp [MAXN]; // dp [I] indicates the total number of solutions for value I
Int n, ans;
 
// Process all data in a table
Void solve (){
Memset (dp, 0, sizeof (dp); dp [0] = 1; // note that dp [0] = 1.
For (int I = 0; I <5; I ++ ){
For (int j = 1; j <MAXN; j ++ ){
If (j-type [I]> = 0) dp [j] + = dp [j-type [I]; // accumulate
}
}
}
Www.2cto.com
Int main (){
// Freopen ("input.txt", "r", stdin );
Solve ();
While (scanf ("% d", & n )! = EOF) printf ("% d \ n", dp [n]);
Return 0;
}

Author: cgl1079743846