zoukankan      html  css  js  c++  java
  • ACM-ICPC 2018 焦作赛区网络预赛 K题 Transport Ship

    There are NN different kinds of transport ships on the port. The i^{th}ith kind of ship can carry the weight of V[i]V[i] and the number of the i^{th}ith kind of ship is 2^{C[i]} - 12C[i]1. How many different schemes there are if you want to use these ships to transport cargo with a total weight of SS?

    It is required that each ship must be full-filled. Two schemes are considered to be the same if they use the same kinds of ships and the same number for each kind.

    Input

    The first line contains an integer T(1 le T le 20)T(1T20), which is the number of test cases.

    For each test case:

    The first line contains two integers: N(1 le N le 20), Q(1 le Q le 10000)N(1N20),Q(1Q10000), representing the number of kinds of ships and the number of queries.

    For the next NN lines, each line contains two integers: V[i](1 le V[i] le 20), C[i](1 le C[i] le 20)V[i](1V[i]20),C[i](1C[i]20), representing the weight the i^{th}ith kind of ship can carry, and the number of the i^{th}ith kind of ship is 2^{C[i]} - 12C[i]1.

    For the next QQ lines, each line contains a single integer: S(1 le S le 10000)S(1S10000), representing the queried weight.

    Output

    For each query, output one line containing a single integer which represents the number of schemes for arranging ships. Since the answer may be very large, output the answer modulo 10000000071000000007.

    样例输入

    1
    1 2
    2 1
    1
    2

    样例输出

    0
    1

    题目来源

    ACM-ICPC 2018 焦作赛区网络预赛

    题解:二进制求和+DP

    参考代码:

     1 #include<iostream>
     2 #include<stdio.h>
     3 #include<math.h>
     4 #include<string.h>
     5 #include<string>
     6 #include<algorithm>
     7 #include<bitset>
     8 #define INF 0x3f3f3f3f
     9 #define clr(x, y) memset(x, y, sizeof(x))
    10 #define mod 1000000007
    11 using namespace std;
    12 typedef long long LL;
    13 const int maxn=25;
    14 const int maxm=10010;
    15 int v[maxn],c[maxn];
    16 LL dp[maxm];
    17 
    18 int main()
    19 {
    20     int t;
    21     scanf("%d", &t);
    22     while(t--)
    23     {
    24         int n, q;
    25         scanf("%d%d", &n, &q);
    26         memset(dp, 0, sizeof(dp));
    27         dp[0] = 1;
    28         for(int i = 1; i <= n; i++)
    29         {
    30             scanf("%d%d", &v[i], &c[i]);
    31             LL cur = 1;
    32             for(int j = 1; j <= c[i]; j++)
    33             {
    34                 for(int k = 10000; k >= cur * v[i]; k--)
    35                     dp[k] = (dp[k] + dp[k - cur * v[i]]) % mod;
    36                 cur<<=1;
    37             }
    38         }
    39         for(int i = 1; i <= q; i++)
    40         {
    41             int s;
    42             scanf("%d", &s);
    43             printf("%lld
    ", dp[s]);
    44         }
    45     }
    46     return 0;
    47 }
    View Code

      

  • 相关阅读:
    Android研究之游戏开发处理按键的响应
    C语言指针的初始化和赋值
    Cloudera CDH 5集群搭建(yum 方式)
    未将对象引用设置到对象的实例--可能出现的问题总结
    内存泄漏以及常见的解决方法
    都能看懂的嵌入式linux/android alsa_aplay alsa_amixer命令行使用方法
    Life is hard!
    EasyUI基础入门之Resiable(可缩放)
    Android -- Looper.prepare()和Looper.loop() —深入版
    vi 命令 使用方法
  • 原文地址:https://www.cnblogs.com/csushl/p/9651985.html
Copyright © 2011-2022 走看看