Coin ChangeTime
limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 14982 Accepted Submission (s): 5070
Problem Descriptionsuppose There is 5 types of coins:50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money.
For example, if we had one cents, then we can make changes with one 10-cent coin and one 1-cent coin, or both 5-cent coins And one 1-cent coin, or one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there is four ways of making changes for one cents with the above coins. Note that we count this there is one of the making change for zero cent.
Write a program to find the total number of different ways of making changes for all amount of money in cents. Your program should is able to handle up to coins.
Inputthe input file contains any number of lines, each one consisting of a number (≤250) for the amount of money in cent S.
Outputfor each input line, output a line containing the number of different ways of making changes with the above 5 types of coins.
Sample Input
1126
Sample Output
413
Authorlily
SOURCE Zhejiang University of Technology network trials because it can only use 100 coins, so the array to add one dimension dp[i][j] represents the sum of I need J coin. DP[I][J] = dp[i-v][j-1]; code;
#include <stdio.h> #include <string.h> #define MAX 300#define COUNT 100const int type[]={50,25,10,5,1}; int DP [MAX] [+]; int max (int a, int b) {return a>b?a:b;} int main () {dp[0][0] = 1, for (int i = 0, i < 5; ++i) {for (int j = Type[i], j <=, ++j) {for (int m = 0; M < 100; ++M) {dp[j][m+1] + = Dp[j-type[i]][m];}}} int n; while (~scanf ("%d", &n)) {int-ans = 0; for (int i = 0; I <=, ++i) ans + dp[n][i];p rintf ("%d\n", ans);} return 0;}
with June
Hdu 2069 Coin Change backpack. Originally intended to use the parent function to write again, found that the code is very similar, did not write