Coins
Time Limit: 3000MS |
| Memory Limit: 30000K |
Total Submissions: 33625 |
| Accepted: 11414 |
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) and he known the price would not more than m.But he didn't know the exact price of the watch.
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 two zeros.
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
多重背包，各种优化的方案贴上。
二进制优化再加上一点小优化（缩小最内层循环的大小）
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 1e5 + 10;
int n, m, l, r, x[maxn], y[maxn], z, dp[maxn];
int main()
{
dp[0] = 1;
while (scanf("%d%d", &n, &m), n + m)
{
l = 1; r = m;
for (int i = 1; i <= m; i++) dp[i] = 0;
for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
for (int i = 1; i <= n; i++) scanf("%d", &y[i]);
for (int i = 1; i <= n; i++)
{
for (int j = 1; y[i]; j <<= 1)
{
z = x[i] * min(j, y[i]); y[i] -= min(y[i], j);
for (int k = min(r, m); k >= max(l, z); k--) dp[k] |= dp[k - z];
}
while (dp[r] && r >= l) --r;
while (dp[l] && l <= r) ++l;
}
int ans = 0;
for (int i = 1; i <= m; i++) ans += dp[i];
printf("%d\n", ans);
}
return 0;
}
另外一种优化方式
#include<queue>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 1e5 + 10;
int n, m, x[maxn], y[maxn], z, dp[maxn], f[maxn], ans;
int main()
{
dp[0] = 1;
while (scanf("%d%d", &n, &m), n + m)
{
for (int i = m; i; i--) ans = dp[i] = 0;
for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
for (int i = 1; i <= n; i++) scanf("%d", &y[i]);
for (int i = 1; i <= n; i++)
{
for (int j = 0; j <= m; j++) f[j] = 0;
for (int j = x[i]; j <= m; j++)
{
if (!dp[j] && dp[j - x[i]] && f[j - x[i]] < y[i])
{
dp[j] = 1; f[j] = f[j - x[i]] + 1;
}
}
}
for (int i = 1; i <= m; i++) ans += dp[i];
printf("%d\n", ans);
}
return 0;
}
单调队列优化
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<bitset>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 50;
int n, m;
int v[maxn], w[maxn];
int l, r, ll, rr;
int a[maxn];
int f[maxn];
int main()
{
while (scanf("%d%d", &n, &m), n + m)
{
for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
for (int i = f[0] = 1; i <= m; i++) f[i] = 0;
for (int i = 1; i <= n; i++)
{
if (w[i] * v[i] >= m)
{
for (int j = v[i]; j <= m; j++) f[j] |= f[j - v[i]];
}
else if (w[i] == 1)
{
for (int j = m; j >= v[i]; j--) f[j] |= f[j - v[i]];
}
else
{
for (int j = 0; j < v[i]; j++)
{
l = 1; r = 0;
for (int k = 0; j + k <= m; k += v[i])
{
if (l <= r && k - a[l] > v[i] * w[i]) l++;
if (f[j + k]) a[++r] = k;
else f[j + k] = l <= r;
}
}
}
}
int ans = 0;
for (int i = 1; i <= m; i++) ans += f[i];
printf("%d\n", ans);
}
return 0;
}
