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

A

Solution: dp+ two points

Dp[i]=max (DP[I],DP[J]+P[I].V) (I>J)

Dp[i] Indicates the maximum value that can be obtained after the establishment of the I point

intn,m;structnode{intL,v;} p[1010];intdp[1010];BOOLCMP (node A,node b) {returnA.L <B.L;}BOOLJudge_oo () {intMax =-1;  for(inti =1; I <= n;i++) Max =Max (MAX,P[I].V); if(Max >= M)return 1; return 0;}BOOLJudge_no () {intsum =0;  for(inti =1; I <= n;i++) sum + =p[i].v; if(Sum >= M)return 0; return 1;}BOOLJUDGE_DP (intL) {memset (DP,0,sizeofDP);  for(inti =1; I <= n;i++){         for(intj =0; J < I; J + +){            if(P[i].l-p[j].l >=L) {Dp[i]= Max (Dp[i],dp[j] +p[i].v); }        }    }    intML =0;  for(inti =1; I <= n;i++) ML =Max (ml,dp[i]); if(ML >= M)return 1; return 0;}intSolveintLintR) {     while(L +1<R) {        intMid = (L + R)/2; if(JUDGE_DP (mid)) L =mid; ElseR =mid; }    returnL;}intMainintargcChar*argv[]) {    //freopen ("data.in", "R", stdin); //freopen ("Data.out", "w", stdout);    intT; CIN>>T;  while(t--) {cin>> N >>M;  for(inti =1; I <= N;i + +) Cin>> P[I].L >>p[i].v; if(Judge_no ()) {puts ("-1"); Continue; }        if(Judge_oo ()) {puts ("oo"); Continue; } sort (P+1, p+n+1, CMP); p[0].L =-100000000, p[0].V =0; intL =1, R =P[N].L; cout<< Solve (l,r) <<Endl; }    return 0;}

B

Solution: Simulation. String emulation

#include <stdio.h>//#include <bits/stdc++.h> #include <string.h> #include <iostream> #include <math.h> #include <sstream> #include <set> #include <queue> #include <map> #include < vector> #include <algorithm> #include <limits.h> #define INF 0x3fffffff#define inf 0x3f3f3f3f#define Lson L,m,rt<<1#define Rson m+1,r,rt<<1|1#define LL long long#define ULL unsigned long longusing namespace Std;int I , J;int n,m;int num_1,num_2;int t;int sum,ans,flag;string s_1,s_2,s_3;string sss;string ss[10000];string c;map<    String,int>q1;map<string,int>q2;int Main () {cin>>t;        while (t--) {cin>>n>>c;        num_1=0;            for (i=0; i<n; i++) {cin>>sss;            Ans=sss.find (".");            S_1=sss.substr (0,ans);            S_2=sss.substr (Ans+1,sss.length ());                if (s_2==c) {int ans_2=sss.find (");      int Ans_3=sss.find (")");          if ((Ans_2!=-1&&ans_2<ans_3)) {//num_1++;                    Ss[num_1++]=sss.substr (0,ans_2);                Cout<<sss.substr (0,ans_2) <<endl; } else if (ans_2==-1| |                    ANS_3==-1) {//num_1++;                    Ss[num_1++]=sss.substr (0,ans);                Cout<<sss.substr (0,ans) <<endl; }}} for (i=0; i<num_1; i++) {//cout<<ss[i]<< "A" <<e        Ndl        }//cout<<num_1<<endl;        for (i=0; i<num_1; i++) {q1[ss[i]]++;                } for (i=0; i<num_1; i++) {if (q2[ss[i]]==0) {q2[ss[i]]=1; if (q1[ss[i]]==1) {cout<<ss[i]<< "."                <<c<<endl;  } else {                  cout<<ss[i]<< "."                <<c<< "" <<q1[ss[i]]<<endl;        }}} q1.clear ();    Q2.clear (); } return 0;}

C

Solution: Tree-like array looking for reverse-order + preprocessing

#include <cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>using namespacestd;Const intM =200010;Long LongAns[m +Ten];intG[m +Ten],c[m +Ten];intLowbit (intx) {    returnX & (-x);}voidInsert (intXintFD) {     while(x<=M) {C[x]+=D; X+=lowbit (x); }}intSum (intx) {    intsum=0;  while(x>0) {sum+=C[x]; X-=lowbit (x); }    returnsum;}voidinit () {memset (G,-1,sizeofG); memset (ans,0,sizeofans); Memset (c,0,sizeofc); g[1] =1;  for(inti =2; I <=200100; i++){        if(G[i] = =-1){             for(intj = I;j <=200100; J + =i) {                if(G[j] = =-1) {G[j]=i; }            }        }    }     for(intI=1; i<=200005; i++) {Insert (g[i],1); Ans[i]= ans[i-1] + (I-Sum (G[i])); }}intMain () {init (); intT; CIN>>T;  while(t--)    {        intN; CIN>>N; cout<<ans[n]<<Endl; }    return 0;}

D

Solution: Template problem, minimum cover circle

#include <iostream>#include<cmath>#include<cstdio>#include<algorithm>using namespacestd;Const Doubleeps=1e-8;structpoint{Doublex, y;} p[505];DoubleDisConstPoint &a,ConstPoint &b) {    returnsqrt ((a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (a.y-b.y));} Point Circumcenter (ConstPoint &a,ConstPoint &b,ConstPoint &c) { //returns the circumcenter of a trianglePoint ret; Doublea1=b.x-a.x,b1=b.y-a.y,c1= (A1*A1+B1*B1)/2; DoubleA2=c.x-a.x,b2=c.y-a.y,c2= (A2*A2+B2*B2)/2; Doubled=a1*b2-a2*B1; Ret.x=a.x+ (C1*B2-C2*B1)/D; Ret.y=a.y+ (A1*C2-A2*C1)/D; returnret;}voidMin_cover_circle (Point *p,intN,point &c,Double&r) {//c is the center and R is the radiusRandom_shuffle (P,p+n);//c=p[0]; R=0;  for(intI=1; i<n;i++)    {        if(Dis (p[i],c) >r+eps)//First Point{C=p[i]; R=0;  for(intj=0; j<i;j++)                if(Dis (p[j],c) >r+eps)//a second point{c.x= (p[i].x+p[j].x)/2; C.y= (P[I].Y+P[J].Y)/2; R=dis (p[j],c);  for(intk=0; k<j;k++)                        if(Dis (p[k],c) >r+eps)//a third point{//Ask Circumscribed Circle Center, three points must not collinearC=circumcenter (P[i],p[j],p[k]); R=dis (p[i],c); }                }        }    }}intMain () {intN;    Point C; DoubleR; intT; CIN>>T;  while(t--) {cin>>N;  for(intI=0; i<n; i++) scanf ("%LF%LF",&p[i].x,&p[i].y);        Min_cover_circle (P,N,C,R); printf ("%.1f%.1f\n", C.X,C.Y); }     return 0;}

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