例题5-10 UVa207-PGA Tour Prize Money（WA）

这道题真的是搞心态的神仙题了，由于到最后还是晕晕乎乎+WA，所以就不多说了，直接贴代码。

题目链接：UVa 207

WA代码：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define DQ 99999
#define eps 1e-8struct Player{char name[30];  //姓名int RD[4];   //四轮成绩int RD_2;   //前两轮成绩int RD_all;   //总成绩int places;   //名次double prize;   //奖金bool is_pro;   //是否为职业运动员bool is_T;   //是否并列
}players[1000];bool cmp1(const Player& a, const Player& b) {   //cmp1和cmp2都保证满轮数，所以不用考虑轮数比较if (a.RD_2 != b.RD_2)return a.RD_2 < b.RD_2;return strcmp(a.name, b.name) < 0;
}
bool cmp2(const Player& a, const Player& b) {if (a.RD_all != b.RD_all)return a.RD_all < b.RD_all;return strcmp(a.name, b.name) < 0;
}
bool cmp3(const Player& a, const Player& b) {   //为末尾的DQ选手排序int s1 = 0, s2 = 0;for(int i=0;i<4;i++)if (a.RD[i] == DQ) {s1 = i;break;}for(int i=0;i<4;i++)if (b.RD[i] == DQ) {s2 = i;break;}if (s1 != s2)return s1 > s2;if (a.RD_all != b.RD_all)return a.RD_all < b.RD_all;return strcmp(a.name, b.name) < 0;
}
int len(int n) {int sum = 0;while (n > 0) {n /= 10;sum++;}return sum;
}int main() {int T, num;double money, prop[100];while (T--) {memset(players, 0, sizeof(players));memset(prop, 0, sizeof(prop));cin >> money;for (int i = 0; i < 70; i++)cin >> prop[i];cin >> num;getchar();for (int i = 0; i < num; i++) {fgets(players[i].name, 20, stdin);if (!strchr(players[i].name, '*'))   //strchr用以在一个串中查找给定字符的第一个匹配处，找到则返回1players[i].is_pro = true;for (int j = 0; j < 4; j++) {int flag = 0;if (!scanf("%d",&players[i].RD[j])) {players[i].RD[j] = DQ;flag = 1;}if (j < 2)players[i].RD_2 += players[i].RD[j];players[i].RD_all += players[i].RD[j];if (flag)break;}char ch[45];fgets(ch, 40, stdin);}sort(players, players + num, cmp1);int pos = 0, pos2 = 0;while (pos < min(num, 70) && players[pos].RD_2 < DQ)  //将pos移动到最后一个没有犯规的选手的后一位上pos++;while (pos < num && players[pos].RD_2 == players[pos - 1].RD_2 && players[pos].RD_2 < DQ)  //在num>70且第70位选手未犯规的情况下，将pos移动到与第70位选手并列的最后一位选手的后一位上pos++;sort(players,players+pos,cmp2);while (pos2<num && pos2 <= pos && players[pos2].RD_all < DQ)  //将pos2移动到最后一个在后两轮没有犯规的选手的后一位上或pos（晋级的最后一位选手的后一位）处pos2++;while (pos2 < num && players[pos2].RD_all == players[pos2 - 1].RD_all && players[pos2].RD_all < DQ)pos2++;if (pos2 != pos)sort(players + min(pos,pos2), players + max(pos,pos2), cmp3);int rank = 1, cur = 0, pos3, cont = 0;while (cur < min(pos,pos2)) {int sum = 0;double sum_mon = 0;for (pos3 = cur; players[pos3].RD_all == players[cur].RD_all; pos3++) {if (players[pos3].is_pro) {sum++;sum_mon += prop[cont++];}}if (sum)sum_mon /= sum;elsesum_mon /= 100;for (int i = cur; i < pos3; i++) {players[i].places = rank;if (players[i].is_pro && sum) {players[i].prize = sum_mon * money / 100.0 + eps;if (sum > 1 && cont-sum<70)players[i].is_T = true;}if (cont - sum >= 70)players[i].is_pro = false;}int numb = pos3 - cur;rank += numb;cur += numb;}cout << "Player Name          Place     RD1  RD2  RD3  RD4  TOTAL     Money Won" << endl;cout << "-----------------------------------------------------------------------" << endl;for (int i = 0; i < pos; i++) {printf("%-20s", players[i].name);  //输出name，不足二十位则在后面用空格填充int N = 10;if (players[i].RD_all < DQ) {cout << players[i].places;N -= len(players[i].places);}if (players[i].is_T) {cout << "T";N--;}for (int i = 0; i < N; i++)   //???cout << " ";N = 4;for (int j = 0; j < 4; j++) {if (players[i].RD[j] != DQ)printf("%-6d", players[i].RD[j]);else {N -= j;break;}}if (N == 4)N = 0;for (int i = 0; i < N; i++)   //???cout << "   ";if (N) {cout << "DQ" << endl;continue;}if (players[i].is_pro) {printf("%-8d", players[i].RD_all); printf("$%9.2lf", players[i].prize);}elseprintf("%d", players[i].RD_all);cout << endl;}if (T)cout << endl;}return 0;
}

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