2015年江西理工大学C语言程序设计竞赛（初级组）

JankTao相亲记

解法：排序

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#include<stdio.h> #include<string.h> #include<iostream> #include<math.h> #include<map> #include<set> #include<vector> #include<algorithm> using namespace std; const double INF = 1e20; const double pi = acos(-1.); int main() {     int t;     int a[100];     while(cin>>t)     {         while(t--)         {         for(int i=0;i<11;i++)         {             cin>>a[i];         }         sort(a,a+11);         printf("%d\n",a[9]);         }     }     return 0; }

David的沙漠之路

解法：贪心，尽量往最大的距离走，加上它经历过的加油站中最大的那个，模拟这个过程需要考虑很多情况

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#include <iostream> #include <stdio.h> #include <queue> using namespace std; struct cmp {     bool operator ()(int &a,int &b)     {         return a<b;     } }; struct node {     int x,y; } e[1005]; int main() {     priority_queue<int,vector<int>,cmp>que;     int T;     scanf("%d",&T);     while(T--)     {         int n,l,s,x,y,cnt=0,flag=0,f=0;         scanf("%d%d%d",&n,&l,&s);         for(int i=0; i<n; i++)         {             scanf("%d%d",&e[i].x,&e[i].y);         }         e[n].x=0,e[n].y=l;         for(int i=0; i<=n; i++,f=0)         {             x=e[i].x,y=e[i].y;             if(y<=s)             {                 que.push(x);                 f=1;                 //最长可以横跨多少加油站，把他们放在队列里             }             while(y>=s)             {                 if(s>=l) break;                 if(!f && s==y)                 {                     que.push(x);                     f=1;                     //没有横跨，刚刚好到达                 }                 if(que.empty())                 {                     printf("TJ\n");                     flag=1;                     break;                 }                 int xx=que.top();//取最近的加油站                 que.pop();                 s+=xx;                 cnt++;             }             if(!f) que.push(x);             if(s>=l)             {                 printf("%d\n",cnt);                 flag=1;                 break;             }             if(flag) break;         }         if(!flag) printf("TJ\n");         while(!que.empty()) que.pop();     }     return 0; }

TJ的文件系统

解法：字符串处理

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#include<stdio.h> //#include<bits/stdc++.h> #include<string.h> #include<iostream> #include<math.h> #include<sstream> #include<set> #include<queue> #include<vector> #include<algorithm> #include<limits.h> #define inf 0x3fffffff #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define LL long long using namespace std; const int maxn = 22; int main() {     int t;     int n;     int i;     int j;     string s,ss;     while(cin>>t)     {         while(t--)         {             string sss="";             int ans;             cin>>n>>s;             for(i=0;i<n;i++)             {                 string sss="";                 cin>>ss;                 ans=ss.find(".");              //   cout<<ans<<endl;                 for(j=ans+1;j<ss.length();j++)                 {                     sss+=ss[j];                 }            //     cout<<sss<<endl;                 if(sss==s)                 {                     cout<<ss<<endl;                 }             }         }     }        return 0; }

握手定理

解法：水题

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#include <stdio.h> int main() {     int SEQ, ACK1, ACK2, T;     scanf("%d", &T);     while(T--)     {         scanf("%d %d %d", &SEQ, &ACK1, &ACK2);         if((ACK1 == SEQ + 1) && (ACK2 == ACK1 + 1))         {             printf("QWN3213\n");         }         else             printf("TJ\n");     }     return 0; }

Alice AND Bob

解法：模拟

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#include<stdio.h> //#include<bits/stdc++.h> #include<string.h> #include<iostream> #include<math.h> #include<sstream> #include<set> #include<queue> #include<vector> #include<algorithm> #include<limits.h> #define inf 0x3fffffff #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define LL long long using namespace std; int a[100000],b[100000],c[100000]; int main() {     int t;     int i,j;     int n,m;     int x,y;     int k;     cin>>t;     while(t--)     {         k=0;         cin>>n>>m;         memset(a,0,sizeof(a));         memset(b,0,sizeof(b));         for(i=0;i<n;i++)         {             cin>>x;             a[x]++;         }         for(i=0;i<m;i++)         {             cin>>y;             b[y]++;         }        for(i=0;i<100000;i++)        {             if(a[i]&&b[i])                 c[k++]=i;        }         for(i=0;i<k;i++)         {             if(i==0)             {               printf("%d",c[i]);             }             else             {                 printf(" %d",c[i]);             }         }         cout<<endl;     }     return 0; }

建群数据

解法：字符串处理时间

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#include<stdio.h> //#include<bits/stdc++.h> #include<string.h> #include<iostream> #include<math.h> #include<sstream> #include<set> #include<queue> #include<vector> #include<algorithm> #include<limits.h> #define inf 0x3fffffff #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define LL long long using namespace std; struct P {     int number;     int hhmmss; }hehe[100000]; int cmp(P a,P b) {     if(a.number==b.number)         return a.hhmmss<b.hhmmss;     else         return a.number<b.number; } string s,ss,sss,ssss; string sssss; int t,n; int a; int i,j; int ans; int poi; int num; int main() {        cin>>t;     while(t--)     {         poi=1;         cin>>n;         for(i=0;i<n;i++)         {             cin>>a>>s;             hehe[i].number=a;             ss=s.substr(0,2);             sss=s.substr(2,2);             ssss=s.substr(4,2);             //   cout<<ss<<" "<<sss<<" "<<ssss<<endl;             hehe[i].hhmmss=((ss[0]-'0')*10+(ss[1]-'0')*1)*3600+((sss[0]-'0')*10+(sss[1]-'0'))*60+((ssss[0]-'0')*10+(ssss[1]-'0'));         }         cin>>sssss;         ss=sssss.substr(0,2);         sss=sssss.substr(2,2);         ssss=sssss.substr(4,2);         ans=((ss[0]-'0')*10+(ss[1]-'0'))*3600+((sss[0]-'0')*10+(sss[1]-'0'))*60+((ssss[0]-'0')*10+(ssss[1]-'0'));         sort(hehe,hehe+n,cmp);         num=hehe[0].hhmmss;         for(int i=1;i<n;i++)         {             if(hehe[i].number!= hehe[i-1].number||hehe[i].hhmmss-num>ans)             {                 poi++;                 num=hehe[i].hhmmss;             }         }         printf("%d\n",poi);     }     return 0; }