pat 1062. Talent and Virtue (25)

来源：互联网发布：二级vb语言程序设计编辑：程序博客网时间：2024/06/06 02:45

难得的一次ac

题目意思直接，方法就是对virtue talent得分进行判断其归属类型，用0 1 2 3 4 表示不合格 sage noblemen foolmen foolmen

再对序列进行排序优先级类型>total grade>virtue>id

#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define N 100005struct node{char id[10];int v,t,c;}s[N];int cmp(const void *a,const void *b){struct node *x=(struct node *)a;struct node *y=(struct node *)b;if(x->c!=y->c)return x->c-y->c;if(y->v+y->t!=(x->v+x->t))return y->v+y->t-(x->v+x->t);if(y->v!=x->v)return y->v-x->v;return strcmp(x->id,y->id);}int main(){int n,i,l,h;while(scanf("%d%d%d",&n,&l,&h)!=EOF){for(i=0;i<n;i++){scanf("%s%d%d",s[i].id,&s[i].v,&s[i].t);if(s[i].v>=h&&s[i].t>=h)s[i].c=1;elseif(s[i].v>=h&&s[i].t>=l&&s[i].t<h)s[i].c=2;elseif(s[i].v>=l&&s[i].v<h&&s[i].t>=l&&s[i].t<h&&s[i].v>=s[i].t)s[i].c=3;elseif(s[i].v>=l&&s[i].t>=l)s[i].c=4;elses[i].c=0;}qsort(s,n,sizeof(struct node),cmp);for(i=0;i<n;i++)if(s[i].c!=0)break;printf("%d\n",n-i);for(;i<n;i++)printf("%s %d %d\n",s[i].id,s[i].v,s[i].t);}}