{"id":1445,"date":"2013-04-03T02:08:56","date_gmt":"2013-04-02T18:08:56","guid":{"rendered":"https:\/\/blog.rexdf.org\/?p=1445"},"modified":"2013-04-03T02:08:56","modified_gmt":"2013-04-02T18:08:56","slug":"codeforces-177-d","status":"publish","type":"post","link":"https:\/\/blog.rexdf.org\/ja\/2013\/04\/codeforces-177-d\/","title":{"rendered":"Codeforces 177 D"},"content":{"rendered":"
\n
Polo the Penguin and Houses<\/div>\n
\n
time limit per test<\/div>\n

2 seconds<\/p><\/div>\n

\n
memory limit per test<\/div>\n

256 megabytes<\/p><\/div>\n

\n
input<\/div>\n

standard input<\/p><\/div>\n

\n
output<\/div>\n

standard output<\/p><\/div>\n<\/div>\n

\n

Little penguin Polo loves his home village. The village has\u00a0n<\/i>\u00a0houses, indexed by integers from 1 to\u00a0n<\/i>. Each house has a plaque containing an integer, the\u00a0i<\/i>-th house has a plaque containing integer\u00a0p<\/i>i<\/i><\/sub>\u00a0(1\u2009\u2264\u2009p<\/i>i<\/i><\/sub>\u2009\u2264\u2009n<\/i>).<\/p>\n

Little penguin Polo loves walking around this village. The walk looks like that. First he stands by a house number\u00a0x<\/i>. Then he goes to the house whose number is written on the plaque of house\u00a0x<\/i>\u00a0(that is, to house\u00a0p<\/i>x<\/i><\/sub>), then he goes to the house whose number is written on the plaque of house\u00a0p<\/i>x<\/i><\/sub>\u00a0(that is, to house\u00a0p<\/i>p<\/i>x<\/i><\/sub><\/sub>), and so on.<\/p>\n

We know that:<\/p>\n

 <\/p>\n

    \n
  1. When the penguin starts walking from any house indexed from 1 to\u00a0k<\/i>, inclusive, he can walk to house number 1.<\/li>\n
  2. When the penguin starts walking from any house indexed from\u00a0k<\/i>\u2009+\u20091\u00a0to\u00a0n<\/i>, inclusive, he definitely cannot walk to house number 1.<\/li>\n
  3. When the penguin starts walking from house number 1, he can get back to house number 1 after some non-zero number of walks from a house to a house.<\/li>\n<\/ol>\n

     <\/p>\n

    You need to find the number of ways you may write the numbers on the houses’ plaques so as to fulfill the three above described conditions. Print the remainder after dividing this number by\u00a01000000007\u00a0(109<\/sup>\u2009+\u20097).<\/p>\n<\/div>\n

    \n
    Input<\/div>\n

    The single line contains two space-separated integers\u00a0n<\/i>\u00a0and\u00a0k<\/i>\u00a0(1\u2009\u2264\u2009n<\/i>\u2009\u2264\u20091000,\u20091\u2009\u2264\u2009k<\/i>\u2009\u2264\u2009min<\/i>(8,\u2009n<\/i>)) \u2014 the number of the houses and the number\u00a0k<\/i>\u00a0from the statement.<\/p>\n<\/div>\n

    \n
    Output<\/div>\n

    In a single line print a single integer \u2014 the answer to the problem modulo\u00a01000000007\u00a0(109<\/sup>\u2009+\u20097).<\/p>\n<\/div>\n

    \n
    Sample test(s)<\/div>\n
    \n
    \n
    input<\/div>\n
    5 2<\/pre>\n<\/div>\n
    \n
    output<\/div>\n
    54<\/pre>\n<\/div>\n
    \n
    input<\/div>\n
    7 4<\/pre>\n<\/div>\n
    \n
    output<\/div>\n
    1728<\/pre>\n<\/div>\n<\/div>\n<\/div>\n
    \r\n#include <iostream>\r\n\r\nusing namespace std;\r\n\r\nconst long long MOD=1e9+7;\r\n\r\nlong long pow_mod(int n,int k)\r\n{\r\n    long  long ret=1,p=n;\r\n    while(k)\r\n    {\r\n        if(k&1)\r\n        {\r\n            ret*=p;\r\n            ret%=MOD;\r\n        }\r\n        p*=p;\r\n        p%=MOD;\r\n        k>>=1;\r\n    }\r\n    return ret;\r\n}\r\n\r\nint main()\r\n{\r\n    int N,K;\r\n    cin>>N>>K;\r\n    cout<<((pow_mod(K,K-1)*pow_mod(N-K,N-K))%MOD)<<endl;\r\n    return 0;\r\n}\r\n<\/pre>\n","protected":false},"excerpt":{"rendered":"
    Polo the Penguin and Houses time limit p … \u7d9a\u304d\u3092\u8aad\u3080 →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[3],"tags":[125],"class_list":["post-1445","post","type-post","status-publish","format-standard","hentry","category-algorithm","tag-codeforces"],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/blog.rexdf.org\/ja\/wp-json\/wp\/v2\/posts\/1445","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.rexdf.org\/ja\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.rexdf.org\/ja\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.rexdf.org\/ja\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.rexdf.org\/ja\/wp-json\/wp\/v2\/comments?post=1445"}],"version-history":[{"count":0,"href":"https:\/\/blog.rexdf.org\/ja\/wp-json\/wp\/v2\/posts\/1445\/revisions"}],"wp:attachment":[{"href":"https:\/\/blog.rexdf.org\/ja\/wp-json\/wp\/v2\/media?parent=1445"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.rexdf.org\/ja\/wp-json\/wp\/v2\/categories?post=1445"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.rexdf.org\/ja\/wp-json\/wp\/v2\/tags?post=1445"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}