POJ 1742 Coins
来源:互联网 发布:程序员等级考试 编辑:程序博客网 时间:2024/06/06 07:15
Coins
Time Limit: 3000MS Memory Limit: 30000KTotal 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.
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
3 101 2 4 2 1 12 51 4 2 10 0
Sample Output
84
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;}
0 0
- poj 1742 Coins
- POJ 1742 Coins
- POJ 1742 Coins
- poj 1742 Coins
- poj 1742 Coins
- POJ 1742 coins
- POJ 1742 Coins
- poj 1742 Coins
- POJ 1742 Coins
- poj 1742 coins
- POJ 1742 Coins (DP)
- poj 1742 Coins
- poj 1742 Coins(dp)
- POJ 1742 Coins
- poj 1742 Coins
- 【POJ 1742】Coins
- poj 1742 Coins
- POJ 1742 Coins
- <<web>>2D,transform
- 算法05 之红-黑树
- 2016年武汉科技大学邀请赛网络赛 B题
- IOS9 那些坑
- DNN、CNN、RNN简析
- POJ 1742 Coins
- hdu 4044 GeoDefense 树形DP+分组背包
- JVM -XX: 参数介绍
- java中文乱码解决之道(七)—–JSP页面编码过程
- Java的可变类与不可变类
- [wpf]如果根据其他控件改变按钮的Path样式
- python核心编程第二版中的网络爬爬虫修改
- vi里怎样批量缩进
- 不重复随机数列生成算法2-数组有效位置的最后一个元素移动到当前位置