• hdu 4541 Ten Googol


    http://acm.hdu.edu.cn/showproblem.php?pid=4541

    打表找规律

     1 #include <cstdio>
     2 #include <cstring>
     3 #include <algorithm>
     4 using namespace std;
     5 
     6 int t;
     7 int n,m;
     8 int a[25][25]={{4, 1, 2, 6, 10},{4, 0, 4, 5, 9},{6, 3, 7, 8, 40, 50, 60},{6, 11, 12, 20, 30, 80, 90},{3, 15, 16, 70}, {13,13, 14 ,18, 19 ,41 ,42 ,46 ,51 ,52 ,56 ,61 ,62 ,66}, {26,17 ,21 ,22 ,26 ,31 ,32 ,36 ,44 ,45 ,49 ,54 ,55 ,59 ,64 ,65 ,69 ,81 ,82 ,86 ,91 ,92 ,96, 101}};
     9 
    10 
    11 int main()
    12 {
    13     scanf("%d",&t);
    14     int cas=1;
    15     while(t--)
    16     {
    17         scanf("%d%d",&n,&m);
    18         printf("Case #%d: ",cas++);
    19         if(m>a[n-3][0])
    20         {
    21             printf("-1
    ");
    22         }
    23         else if(n==9&&m==23)
    24         {
    25             printf("10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
    ");
    26         }
    27         else if(n==9&&m==24)
    28         {
    29              printf("20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
    ");
    30         }
    31         else if(n==9&&m==25)
    32         {
    33              printf("60000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
    ");
    34         }
    35         else if(n==9&&m==26)
    36         {
    37             printf("100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
    ");
    38         }
    39         else
    40         {
    41             printf("%d
    ",a[n-3][m]);
    42         }
    43     }
    44     return 0;
    45 }
    View Code
  • 相关阅读:
    poj 1584
    poj 1113 & poj 2187
    pku 1321 棋盘问题
    poj 1408
    pku 2251 Dungeon Master
    sdut oj 2218 Give Me an E
    Android工程 单元测试
    Android Timer编写方式
    去除工程的.svn隐藏文件夹
    Android 绑定远程服务出现 Not Allowed to bind service
  • 原文地址:https://www.cnblogs.com/fanminghui/p/4024966.html
Copyright © 2020-2023  润新知