2015 C Language Program design Competition (junior group), Jiangxi University of Technology

Janktao Blind Date

Solution: Sort

#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's path to the desert

Solution: Greedy, as far as possible to go to the maximum distance, plus it experienced the largest gas station, simulation of this process need to consider a lot of circumstances

#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;                The longest can span the number of gas stations, put them in the queue} while (Y>=s) {if (s>=l) break;                    if (!f && s==y) {que.push (x);                    f=1;             Not across, just good to arrive}   if (Que.empty ()) {printf ("tj\n");                    flag=1;                Break                } int xx=que.top ();//Take the nearest gas station 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's File system

Solution: String Processing

#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 longusing 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;}

Handshake theorem

Solution: Water Problem

#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

Solution: Simulation

#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 longusing 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;}

Building Group Data

Solution: String Processing time

#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 longusing 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 ') * * *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;}

　　

2015 C Language Program design Competition (junior group), Jiangxi University of Technology