codeforces493 A. Vasya and Football B. Vasya and Wrestling C. Vasya and Basketball

 A. Vasya and Football

这个题目很简单，只需要按照提议模拟一下即可，注意模拟的时候要先按照时间排序，一旦某个人已经被提示红牌，就不在对其考虑；

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define MAX 110#define ll __int64using namespace std;char sname[30], tname[30];int a[MAX],b[MAX];struct Node{int time;      //被警告的时间 char c1[10];        //所属的队名 int num;char c2[10];       //是什么类型的警告 bool operator < (const Node& rhs ) const{return rhs.time > time;}}p[MAX];    //记录每个队员的属性 int main(){int n;while (scanf("%s%s",sname,tname) != EOF){scanf("%d",&n);for (int i = 0; i<n; i++){scanf("%d%s%d%s",&p[i].time,p[i].c1,&p[i].num,p[i].c2);}sort(p,p+n);memset(a,0,sizeof(a));     //记录'h'队的队员是否已经被提示红牌 memset(b,0,sizeof(b));     //记录'a'队的队员是否已经被提示红牌 //for (int i = 0; i<n; i++) printf("%d ",p[i].time); printf("\n");for (int i = 0; i<n; i++){char u = p[i].c1[0], v = p[i].c2[0];int x = p[i].time, y = p[i].num;if (u == 'h'){if (a[y] >= 2){   //该队员若已经被提示红牌，则忽略 continue;}if (v == 'r'){printf("%s %d %d\n",sname,y,x);a[y] = 2;}if (v == 'y'){a[y]++;if (a[y] >= 2){printf("%s %d %d\n",sname,y,x);}}}else{if (b[y] >= 2){continue;}if (v == 'r'){printf("%s %d %d\n",tname,y,x);b[y] = 2;}if (v == 'y'){b[y]++;if (b[y] >= 2){printf("%s %d %d\n",tname,y,x);}}}}}}

 B. Vasya and Wrestling 

很水的模拟题，只需要按照题目的条件逐一判断即可，一旦在某个条件下能够分出胜负，就不在判断,注意可能会爆int：

判断的顺序为：

1、和是否相同；

2、字典序是否相同；

3、最后一次后奖励的是谁；

#include<cstdio>#include<cstring>#include<algorithm>#define MAX 100010#define ll __int64using namespace std;ll a[MAX],b[MAX];int main(){int n;int u,v;while (scanf("%d",&n) != EOF){u = 0;v = 0;ll t;ll sum1 = 0, sum2 = 0;for (int i = 0; i<n; i++){scanf("%I64d",&t);if (t > 0){a[u++] = t;sum1 += t;}else{b[v++] = -t;sum2 += (-t);}}if (sum1 > sum2){printf("first\n");continue;}if (sum1 < sum2){printf("second\n");continue;}//for (int i = 0; i<u; i++) printf("%d ",a[i]); printf("\n");//for (int i = 0; i<v; i++) printd("%d ",b[i]); printf("\n");int sign = -1;for (int i = 0; i<min(u,v); i++){if (a[i] > b[i]){sign = 0;break;}if (a[i] < b[i]){sign = 1;break;}}if (sign == 0){ printf("first\n"); continue;}if (sign == 1){printf("second\n");continue;}if (u > v){ printf("first\n"); continue;}if (u < v){printf("second\n");continue;}if (t > 0){printf("first\n");}else{printf("second\n");}}return 0;}

C. Vasya and Basketball

这个题目当时做的时候想了一会，觉得是用二分，并且c++里面有很好的函数库可以用，就开始写了，结果脑残了一下，最后有个条件(a应该取最大值)没有判断，WA了两次，耽误了好久的时间。思路很简单，就是枚举d值，然后二分去找大于d的第一个数的位置，然后直接根据题目要求，找出结果就好，注意，d取0和取无穷大的情况；

#include<cstdio>#include<cstring>#include<algorithm>#define MAX 200010#define ll __int64#define INF 10000000000using namespace std;ll a[MAX],b[MAX];ll c[MAX*2];   //存放所有可能的d值，容易看出d只需取a,b中出现的值和0,INF即可代表所有情况 int main(){    int n,m;    while (scanf("%d",&n) != EOF){        int cnt = 0;        c[cnt++] = 0;        for (int i = 1; i<=n; i++){            scanf("%I64d",&a[i]);            c[cnt++] = a[i];        }                scanf("%d",&m);        for (int i = 1; i<=m; i++){            scanf("%I64d",&b[i]);            c[cnt++] = b[i];        }        c[cnt++] = INF;             sort(a+1,a+n+1);        sort(b+1,b+m+1);        sort(c,c+cnt);        cnt = unique(c,c+cnt) - c;                ll ans = -INF, u, v, t1, t2;        for (int i = 0; i<cnt; i++){            int x = upper_bound(a+1,a+n+1,c[i]) - a;   //找到在a中第一个严格大于c[i]的元素的位置             int y = upper_bound(b+1,b+m+1,c[i]) - b;                     t1 = (x-1)*2 + (n-x+1)*3;            t2 = (y-1)*2 + (m-y+1)*3;            if (t1 - t2 > ans){                ans = t1 - t2;                u = t1;                v = t2;            }            else if (t1 - t2 == ans && t1 > u){   //注意u应该取最大值，因为少加了t1 > u的判断WA了两次 .o(╯□╰)o                 u = t1;                v = t2;            }        }                printf("%I64d:%I64d\n",u,v);            }    return 0;}