您的位置:首页 > 其它

uva 10194 - Football (aka Soccer)

2013-08-15 11:15 447 查看
这道题题意很好明白    就是提取信息 然后进行排序    注意   getchar吃掉scanf的回车    否则就读到gets中去了      还有一点要注意    最后字典序的时候  忽略大小写进行排序  

这种题考的就是细心

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
using namespace std;

struct TEAM
{
char name[150];
int b,c,d,e,f,g,h,i;
}team[50];

int cmp(struct TEAM x, struct TEAM y)
{
if(x.b > y.b)return 1;
else if(x.b < y.b)return 0;
if(x.d > y.d)return 1;
else if(x.d < y.d)return 0;
if(x.g > y.g)return 1;
else if(x.g < y.g)return 0;
if(x.h > y.h)return 1;
else if(x.h < y.h)return 0;
if(x.c < y.c)return 1;
else if(x.c > y.c)return 0;
int i;
for(i = 0 ; x.name[i]!='\0'&&y.name[i]!='\0'; i++)//不考虑大小写
if(tolower(x.name[i]) < tolower(y.name[i]))return 1;
else if(tolower(x.name[i]) > tolower(y.name[i]))return 0;
if(x.name[i] == '\0')return 1;
else return 0;
}

int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif // LOCAL
int T,i,j,k,p,q;
scanf("%d",&T);
getchar();
while(T--)
{
char game[150];
gets(game);
puts(game);
int N;
scanf("%d",&N);
getchar();
for(i = 0; i < N; i++)
{
gets(team[i].name);
team[i].b = team[i].c = team[i].d = team[i].e = team[i].f = team[i].g = team[i].h = team[i].i = 0;
}
int M;
scanf("%d",&M);
getchar();
char str[300],st[150];
for(i = 0; i < M; i++)
{
gets(str);
int gold_x = 0,gold_y = 0;
for(j = 0; j < strlen(str); j++)
{
int t = 0;
while(str[j] != '#' )st[t++]=str[j++];
st[t] = '\0';
for(p = 0; p < N; p++)if(strcmp(team[p].name,st)==0){team[p].c++;break;}//比赛次数+1
j++;
while(str[j]!='@')gold_x = gold_x*10+(str[j++]-'0');
team[p].h += gold_x;//进球数
j++;
while(str[j]!='#')gold_y = gold_y*10+(str[j++]-'0');
team[p].i +=gold_y;//失球数
j++;
t = 0;
while(str[j] != '\0' )st[t++]=str[j++];
st[t] = '\0';
for(q = 0; q < N; q++)if(strcmp(team[q].name,st)==0){team[q].c++;break;}//比赛次数+1
team[q].h += gold_y;//进球数
team[q].i += gold_x;//失球数
//根据进球数判断本场比赛的得分
if(gold_x > gold_y){team[p].b += 3; team[p].d++; team[q].f++;}
else if(gold_x < gold_y) {team[q].b += 3; team[q].d++; team[p].f++;}
else {team[p].b++; team[q].b++; team[p].e++; team[q].e++;}
}
}
for(k = 0; k < N; k++) // 最后计算goal difference
team[k].g = team[k].h - team[k].i;
sort(team,team+N,cmp);//排序
for(k = 0; k < N; k++)//输出
printf("%d) %s %dp, %dg (%d-%d-%d), %dgd (%d-%d)\n",k+1,team[k].name,
team[k].b,team[k].c,team[k].d,team[k].e,team[k].f,team[k].g,team[k].h,team[k].i);
if(T)printf("\n");
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签:  排序 字符串水题