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Language: Default Coins
Description People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony‘s coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins. Input The input contains several test cases. The first line of each test case contains two integers n(1<=n<=100),m(m<=100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1<=Ai<=100000,1<=Ci<=1000). The last test case is followed by Output For each test case output the answer on a single line. Sample Input 3 10 1 2 4 2 1 1 2 5 1 4 2 1 0 0 Sample Output 8 4 Source |
题意：给n张不同面值的钱，每种面值的钱都有一定数量，问用这些钱能够凑出多少种不同的面值，并且面值要在1~m内。输出种数。
思路：dp[i]表示i面值的钱是否能够凑出来（0或1）。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 100005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
int n,m;
int dp[maxn],num[maxn];
int v[111],c[111];
int main()
{
int i,j;
while (sff(n,m)&&(n+m))
{
mem(dp,0);
FRL(i,0,n)
sf(v[i]); //v[i]存的面值
FRL(i,0,n)
sf(c[i]); //c[i]存的每个面值对应的数量
int ans=0;
dp[0]=1;
FRL(i,0,n)
{
mem(num,0); //num[j]表示凑够面值j时用了多少v[i]面值的钱
FRE(j,v[i],m)
{
if (!dp[j]&&dp[j-v[i]]&&num[j-v[i]]<c[i]) //j面值之前没有凑出来过并且j-v[i]面值的凑出来过，这样就可以增加一张v[i]凑成新面值
{
ans++;
dp[j]=1;
num[j]=num[j-v[i]]+1;
}
}
}
pf("%d\n",ans);
}
return 0;
}