这个题真是快把自己折磨疯了,在学校因为要复习,折腾了10天才AC,结果电脑出了点问题代码数据全部丢失,放假后又写了一遍,结果不断RE一天,唉!
这个题的题意就不说了,紫书上说的很详细了,关键是一点,什么时候分配奖金结束,是等那70个奖金比例全部分配结束后就不分了,哪怕最后一个比例分给了10个并列的,也只分一个,后面就不分了(只要让sum_mon加0就行了!)
整体思路:
1.开一个结构体记录运动员的:4个成绩(dq[5]),姓名(name),奖金(money),名次(place),是否是专业运动员(is_pro),是否输出T(is_T),和前两句的分数总和(all_2),前四局的分数总和(all_4)。
因为输入分数会有字母DQ,不妨设DQ为一个常量99999,这样排序时自动到后面去了
2.输入完成后就是各种排序
3.处理名次和奖金,先找重复的,在给这些重复的定名次和奖金
4.输出结果时,找出各种情况的相同点,来减少代码量!
技巧:
1.将DQ设置为一个很大的常量,让正常分数达不到这个值。
2.读取名字可以用fgets()来读,控制指定的长度(20个字符)。(记得最后在用fgets吃掉换行!)
3.直接用memset给结构体进行初始化为0.(为了简便,将一些改动小的默认为0.例如是否为专业运动员)
4.将比例数值大于70的那些运动员自动设为非职业运动员,这样输出可以直接判断是否为职业运动员来进行输出奖金。
注意:
1.千万注意好访问运动员(结构体)的个数,不要非法访问内存。(RE一天的原因T_T !)
2.在一个奖金平分时会除以个数,注意好别除以0
3.注意奖金要加上精度控制(1e-8就可以了)
代码如下:
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn = 200; const int maxt = 50; const int DQ = 999999; const double eps = 1e-8; struct player { char name[maxt]; double money; int place,all_2,all_4,dq[5]; bool is_pro,is_T; }players[maxn]; int rev_len(int k){ int sum = 0; while(k!=0){ k/=10; sum++; } return sum; } bool cmp1(const player&a,const player&b){ if (a.all_2 != b.all_2)return a.all_2 < b.all_2; return strcmp(a.name,b.name)<0; } bool cmp2(const player&a,const player&b){ if (a.all_4!=b.all_4)return a.all_4<b.all_4; return strcmp(a.name,b.name) < 0; } bool cmp3(const player&a,const player&b){ int suma=0,sumb=0; for (int i = 0; i < 4;++i)if (a.dq[i]==DQ){suma=i;break;} for (int i = 0; i < 4;++i)if (b.dq[i]==DQ){sumb=i;break;} if (suma!=sumb)return suma>sumb; if(a.all_4!=b.all_4)return a.all_4<b.all_4; return strcmp(a.name,b.name)<0; } int main() { //freopen("out.txt","w",stdout); int T,n; double prot[maxn],money_all; scanf("%d",&T);{ while(T--){ memset(players,0,sizeof(players)); memset(prot,0,sizeof(prot)); scanf("%lf",&money_all); for (int i = 0; i < 70; ++i)scanf("%lf",&prot[i]); scanf("%d",&n); getchar(); for (int i = 0; i < n; ++i){ fgets(players[i].name,20,stdin); if (!strchr(players[i].name, '*'))players[i].is_pro=true; for (int j = 0; j < 4; ++j){ int flag = 0; if (!scanf("%d",&players[i].dq[j])){ players[i].dq[j] = DQ; flag = 1; } if (j<2)players[i].all_2+=players[i].dq[j]; players[i].all_4+=players[i].dq[j]; if (flag)break; } char ch[45]; fgets(ch,40,stdin); } sort (players,players+n,cmp1); int pos=0,pos2=0; while(pos < min(n,70) && players[pos].all_2 < DQ)++pos; while(pos < n && players[pos].all_2 == players[pos-1].all_2 && players[pos].all_2 < DQ)++pos; sort(players,players+pos,cmp2); while(pos2 < n && pos2 <= pos && players[pos2].all_4 < DQ)++pos2; while(pos2 < n && players[pos2].all_4 == players[pos2-1].all_4 && players[pos2].all_4<DQ)++pos2; if (pos!=pos2)sort(players+min(pos,pos2),players+max(pos,pos2),cmp3); int rak=1,cur=0,pos3,cont3=0; while(cur < min(pos,pos2) ){ int sum=0; double sum_mon=0; for (pos3 = cur; players[pos3].all_4==players[cur].all_4;++pos3){ if (players[pos3].is_pro){ sum++; sum_mon+=prot[cont3++]; } } if (sum)sum_mon/=sum; else sum_mon/=100; for (int i = cur; i < pos3; ++i){ players[i].place=rak; if (players[i].is_pro && sum)players[i].money=sum_mon*money_all/100.0+eps; if(players[i].is_pro && sum > 1 && cont3-sum<70)players[i].is_T=true; if (cont3-sum >=70)players[i].is_pro=false; } int numb = pos3-cur; rak+=numb; cur+=numb; } printf("Player Name Place RD1 RD2 RD3 RD4 TOTAL Money Won\n"); printf("-----------------------------------------------------------------------\n"); for (int i = 0; i < pos; ++i){ printf("%-21s",players[i].name); int N=10; if (players[i].all_4<DQ){printf("%d",players[i].place);N-=rev_len(players[i].place);} if (players[i].is_T){printf("T");N--;} for (int i = 0; i < N; ++i)printf(" ");N=4; for (int j =0;j < 4; ++j){ if(players[i].dq[j]!=DQ) printf("%-5d",players[i].dq[j]); else {N-=j;break;} }if (N==4)N=0; for(int i = 0; i < N; ++i)printf(" "); if (N){printf("DQ\n");continue;} if (players[i].is_pro){printf("%-10d",players[i].all_4);printf("$%9.2lf",players[i].money);} else printf("%d",players[i].all_4); printf("\n"); } if (T)printf("\n"); } } return 0; }