背包解决硬币问题专题

link：http://acm.hdu.edu.cn/showproblem.php?pid=1284

钱币兑换问题

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6727    Accepted Submission(s): 3903

Problem Description

在一个国家仅有1分，2分，3分硬币，将钱N兑换成硬币有很多种兑法。请你编程序计算出共有多少种兑法。

 

Input

每行只有一个正整数N，N小于32768。

 

Output

对应每个输入，输出兑换方法数。

 

Sample Input

293412553

 

Sample Output

71883113137761

 

Author

SmallBeer(CML)

 

Source

杭电ACM集训队训练赛（VII）

 

AC code：

#include<iostream>#include<stdio.h>#include<map>#include<vector>#include<set>#include<cstdlib>#include<string.h>#include<string>#include<algorithm>#include<cmath>#define MAXN 1000010#define LL long long#define EPS 1e-9using namespace std;int dp[MAXN],c[MAXN];int main(){    int n,i,j;    c[1]=1;    c[2]=2;    c[3]=3;    while(~scanf("%d",&n))    {        memset(dp,0,sizeof(dp));        dp[0]=1;        for(i=1;i<=3;i++)        {            for(j=c[i];j<=n;j++)            {                dp[j]+=dp[j-c[i]];            }        }        printf("%d\n",dp[n]);    }    return 0;}

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1028

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14668    Accepted Submission(s): 10329

Problem Description

"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.

"The second problem is, given an positive integer N, we define an equation like this:
  N=a[1]+a[2]+a[3]+...+a[m];
  a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
  4 = 4;
  4 = 3 + 1;
  4 = 2 + 2;
  4 = 2 + 1 + 1;
  4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"

 

Input

The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.

 

Output

For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

 

Sample Input

41020

 

Sample Output

542627

 

Author

Ignatius.L

 

AC  code：

#include<iostream>#include<stdio.h>#include<map>#include<vector>#include<set>#include<cstdlib>#include<string.h>#include<string>#include<algorithm>#include<cmath>#define MAXN 1000010#define LL long long#define EPS 1e-9using namespace std;int dp[MAXN],c[MAXN];int main(){int n,i,j;for(i=1;i<=122;i++)c[i]=i;while(~scanf("%d",&n)){memset(dp,0,sizeof(dp));dp[0]=1;for(i=1;i<=n;i++){for(j=c[i];j<=n;j++){dp[j]+=dp[j-c[i]];}}printf("%d\n",dp[n]);}return 0;}

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1398

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8983    Accepted Submission(s): 6135

Problem Description

People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ..., and 289-credit coins, are available in Silverland. 
There are four combinations of coins to pay ten credits: 

ten 1-credit coins,
one 4-credit coin and six 1-credit coins,
two 4-credit coins and two 1-credit coins, and
one 9-credit coin and one 1-credit coin. 

Your mission is to count the number of ways to pay a given amount using coins of Silverland.

 

Input

The input consists of lines each containing an integer meaning an amount to be paid, followed by a line containing a zero. You may assume that all the amounts are positive and less than 300.

 

Output

For each of the given amount, one line containing a single integer representing the number of combinations of coins should be output. No other characters should appear in the output. 

 

Sample Input

210300

 

Sample Output

1427

 

Source

Asia 1999, Kyoto (Japan)

 

AC code：

#include<iostream>#include<stdio.h>#include<map>#include<vector>#include<set>#include<cstdlib>#include<string.h>#include<string>#include<algorithm>#include<cmath>#define MAXN 1000010#define LL long long#define EPS 1e-9using namespace std;int dp[MAXN],c[MAXN];int main(){int n,i,j;for(i=1;i<=17;i++)c[i]=i*i;while(~scanf("%d",&n)&&n){memset(dp,0,sizeof(dp));dp[0]=1;for(i=1;i<=17;i++){for(j=c[i];j<=n;j++){dp[j]+=dp[j-c[i]];}}printf("%d\n",dp[n]);}return 0;}

Link：http://acm.hdu.edu.cn/showproblem.php?pid=2069

Coin Change

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14976    Accepted Submission(s): 5066

Problem Description

Suppose there are 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 have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, or two 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 are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent.

Write a program to find the total number of different ways of making changes for any amount of money in cents. Your program should be able to handle up to 100 coins.

 

Input

The input file contains any number of lines, each one consisting of a number ( ≤250 ) for the amount of money in cents.

 

Output

For 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

 

Author

Lily

 

Source

浙江工业大学网络选拔赛

 

注意：此题要求硬币总数不超过100

AC  code：

#include<iostream>#include<stdio.h>#include<map>#include<vector>#include<set>#include<cstdlib>#include<string.h>#include<string>#include<algorithm>#include<cmath>#define MAXN 1000010#define LL long long#define EPS 1e-9using namespace std;int dp[333][333],c[MAXN];LL ans;int main(){int n,i,j,k;c[1]=50;c[2]=25;c[3]=10;c[4]=5;c[5]=1;while(~scanf("%d",&n)){memset(dp,0,sizeof(dp));dp[0][0]=1;for(i=1;i<=5;i++){for(k=0;k<=100;k++){for(j=c[i];j<=n;j++)    {    dp[j][k]+=dp[j-c[i]][k-1];    }}}ans=0;for(i=0;i<=100;i++){ans+=dp[n][i];}printf("%d\n",ans);}return 0;}