zoukankan      html  css  js  c++  java
  • POJ 1742

    Coins
    Time Limit: 3000MS   Memory Limit: 30000K
    Total Submissions: 27580   Accepted: 9335

    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
    

    Source

     
    定义dp[i]为目前为止装满i容量最多剩余多少个当前面值的
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 
     6 using namespace std;
     7 
     8 const int MAX_N = 105;
     9 const int MAX_M = 1e5 + 7;
    10 int N, M;
    11 int v[MAX_N], c[MAX_N];
    12 int dp[MAX_M];
    13 
    14 void solve() {
    15         dp[0] = 0;
    16         for(int i = 1; i <= N; ++i) {
    17                 for(int j = 0; j <= M; ++j) {
    18                         if(dp[j] >= 0) {
    19                                 dp[j] = c[i];
    20                         }
    21                         else if(j < v[i] || dp[j - v[i]] <= 0) {
    22                                 dp[j] = -1;
    23                         } else {
    24                                 dp[j] = dp[j - v[i]] - 1;
    25                         }
    26 
    27                 }
    28         }
    29 
    30 
    31         int ans = 0;
    32         for(int i = 1; i <= M; ++i) {
    33                 ans += dp[i] >= 0;
    34                 //printf("%d ",dp[i]);
    35         }
    36 
    37         printf("%d
    ", ans);
    38 }
    39 int main()
    40 {
    41    // freopen("sw.in", "r", stdin);
    42     while(~scanf("%d%d", &N, &M) && (N + M)) {
    43             memset(dp, -1, sizeof(dp));
    44             for(int i = 1; i <= N; ++i) {
    45                     scanf("%d", &v[i]);
    46             }
    47             for(int i = 1; i <= N; ++i) {
    48                     scanf("%d", &c[i]);
    49             }
    50 
    51             solve();
    52     }
    53     //cout << "Hello world!" << endl;
    54     return 0;
    55 }
    View Code
  • 相关阅读:
    selenium产生的垃圾文件清理
    英语学习词根法
    ecommerce学习
    tfs2012安装
    转载文件,英语学习
    Mishka and Divisors CodeForces
    Codeforces Round #364 (Div. 1) (差一个后缀自动机)
    莫比乌斯反演练习
    bzoj 1267 Kth Number I (点分治,堆)
    程序员的绘图利器 — Gnuplot
  • 原文地址:https://www.cnblogs.com/hyxsolitude/p/3726867.html
Copyright © 2011-2022 走看看