Sicily 4873. D’HONDT

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2012年每周一赛第二场第一题，简单模拟问题。题目的意思是，对于票数大于5%的政党，将其总票数分别从1除到14，得出14个值，然后把每个政党的这些值混在一起排序，那么最大的14个值所属的政党中，将有一人被选举为代表，然后就是按字典序输出每个政党的代表人数。需要注意的是，即使这个政党没人当选为代表，也是要输出的，只要他的得票数大于总票数的5%。

Run Time: 0sec

Run Memory: 304KB

Code Length: 898Bytes

Submit Time: 2012-03-03 21:45:54

// Problem#: 4873// Submission#: 1231763// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <cstdio>#include <map>using namespace std;int main(){    int X, N, G;    char P;    map<char,int> vote, rept;    map<char,int>::iterator cit;    map<double,char> score;    map<double,char>::iterator dit;    int i;    scanf( "%d%d", &X, &N );    for ( i = 1; i <= N; i++ ) {        getchar();        scanf( "%c%d", &P, &G );        if ( (double)G / X >= 0.05 ) {            vote[ P ] = G;            rept[ P ] = 0;        }    }    if ( N != 0 ) {        for ( cit = vote.begin(); cit != vote.end(); cit++ ) {            for ( i = 1; i <= 14; i++ )                score[ (double)cit->second / i ] = cit->first;        }        for ( i = 1, dit = --score.end(); i <= 14; i++, dit-- )            rept[ dit->second ]++;        for ( cit = rept.begin(); cit != rept.end(); cit++ )            printf( "%c %d\n", cit->first, cit->second );    }    return 0;}