PAT甲级考前整理一：https://www.cnblogs.com/jlyg/p/7525244.html，主要讲了131题的易错题及坑点
PAT甲级考前整理二：https://www.cnblogs.com/jlyg/p/10364696.html，主要讲了考前注意以及一些常用算法。
1132题：用字符串接收会毕竟快，使用atoi函数转成数字，注意a*b会超出int32。
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; int main() { #ifdef ONLINE_JUDGE #else freopen("test.txt","r",stdin); #endif int t; scanf("%d",&t); while(t--) { char str[20]; scanf("%s",str); char str1[20],str2[20]; int len = strlen(str); strncpy(str1,str,len/2); str1[len/2]=' '; strcpy(str2,str+len/2); int a = atoi(str1),b=atoi(str2); int c = atoi(str); //a*b超出int32 if(a*b<=0||c%(a*b)) printf("No "); else printf("Yes "); } return 0; }
1133题：因为k是正数，用3个vector存储就ok了，v0表示存储负数，v1表示存储小于等于k的，v2表示存储大于k的。
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; #define N (100010) struct ST { int addr; int val; int next; }; ST sts[N]; int main() { #ifdef ONLINE_JUDGE #else freopen("test.txt","r",stdin); #endif int startaddr,n,k; scanf("%d%d%d",&startaddr,&n,&k); for(int i=0;i<n;++i) { ST st; scanf("%d%d%d",&st.addr,&st.val,&st.next); sts[st.addr] = st; } vector<ST> vs; for(int addr=startaddr;addr!=-1;addr=sts[addr].next) vs.push_back(sts[addr]); vector<ST> vs0/*负数*/,vs1/*比k小或等于的正数*/,vs2/*比k大的正数*/; for(int i=0;i<vs.size();++i) { if(vs[i].val < 0) vs0.push_back(vs[i]); else if(vs[i].val<=k) vs1.push_back(vs[i]); else if(vs[i].val>k) vs2.push_back(vs[i]); } { vs = vs0; for(int i=0;i<vs1.size();++i) vs.push_back(vs1[i]); for(int i=0;i<vs2.size();++i) vs.push_back(vs2[i]); for(int i=0;i<vs.size();++i) { if(i!=vs.size()-1) printf("%05d %d %05d ",vs[i].addr,vs[i].val,vs[i+1].addr); else printf("%05d %d -1 ",vs[i].addr,vs[i].val); } } return 0; }
1134题：题意有点难懂，顶点覆盖，即所有边的（边有两个顶点，任一一个顶点）某个顶点必须是在已给的顶点集合中。如果用遍历会超时，所以要用hash的想法。
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; #define N (10010) struct Edge //记录边的两个端点 { int a; int b; }; int main() { #ifdef ONLINE_JUDGE #else freopen("test.txt","r",stdin); #endif int n,m; vector<Edge> ve; scanf("%d%d",&n,&m); for(int i=0;i<m;++i) { Edge edge; scanf("%d%d",&edge.a,&edge.b); ve.push_back(edge); } int k; scanf("%d",&k); while(k--) { int nv; scanf("%d",&nv); bool flag = true;//是否是顶点覆盖 bool V[N]; memset(V,0,sizeof(V)); //下标表示定点，值表示是否有无改点 for(int i=0;i<nv;++i) { int a; scanf("%d",&a); V[a] = true; } for(int i=0;i<ve.size();++i) { flag = V[ve[i].a]||V[ve[i].b]; if(flag==false) break; } if(flag) printf("Yes "); else printf("No "); } return 0; }
1135题：红黑树的题目，给出先序（插入顺序）构造红黑树，然后判断是否是红黑树。可以看我写的另一篇博客：
https://www.cnblogs.com/jlyg/p/7542409.html
1136题：简单题，大数加法加字符串反转
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; string Add(string str1,string str2) { int len = str1.length(); int add = 0; string res = ""; for(int i=len-1;i>=0;--i) { int k = str1[i]-'0'+str2[i]-'0' + add; add = k/10; res += (k%10+'0'); } if(add) res += (add+'0'); reverse(res.begin(),res.end()); return res; } int main() { #ifdef ONLINE_JUDGE #else freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE char temp[1010]; scanf("%s",temp); string str = temp,str2=temp; for(int i=0;i<10;++i) { reverse(str2.begin(),str2.end()); if(strcmp(str.c_str(),str2.c_str())==0) { printf("%s is a palindromic number. ",str.c_str()); return 0; } string str3 = Add(str,str2); printf("%s + %s = %s ",str.c_str(),str2.c_str(),str3.c_str()); str = str2 = str3; } printf("Not found in 10 iterations. "); return 0; }
1137题：简单题，Gm若是大于Gf,则G=(0.4*Gm+0.6*Gf+0.5)，否则G = Gf; Gp<200和G<60的不记录。
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; struct ST { string id; int Gp; int Gm; int Gf; int G; ST(){Gp=Gm=Gf=G=-1;} void UpdateG() { if(Gm > Gf) G = int(Gm*0.4f+Gf*0.6f+0.5f); else G = Gf; } bool IsUseful() { if(Gp<200) return false; if(G<60) return false; return true; } }; int cmp(const ST& st1,const ST& st2) { if(st1.G != st2.G) return st1.G>st2.G; return strcmp(st1.id.c_str(),st2.id.c_str())<0; } int main() { #ifdef ONLINE_JUDGE #else freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int p,m,n; scanf("%d%d%d",&p,&m,&n); map<string,ST> mss; for(int i=0;i<p;++i) { char strid[30]; int g; scanf("%s%d",strid,&g); mss[strid].Gp = g; mss[strid].id = strid; } for(int i=0;i<m;++i) { char strid[30]; int g; scanf("%s%d",strid,&g); mss[strid].Gm = g; mss[strid].id = strid; } for(int i=0;i<n;++i) { char strid[30]; int g; scanf("%s%d",strid,&g); mss[strid].Gf = g; mss[strid].id = strid; } vector<ST> vs; for(map<string,ST>::iterator it = mss.begin();it!=mss.end();it++) { it->second.UpdateG(); if(it->second.IsUseful()) vs.push_back(it->second); } sort(vs.begin(),vs.end(),cmp); for(int i=0;i<vs.size();++i) { printf("%s %d %d %d %d ",vs[i].id.c_str(),vs[i].Gp,vs[i].Gm,vs[i].Gf,vs[i].G); } return 0; }
1138题：简单题，先序和中序求后序第一个元素的值。
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; int FindFirst(int* pre,int* inorder,int n) { if(n==1) return *pre; int val = *pre; int iIndex; for(iIndex=0;iIndex<n;++iIndex) { if(val == inorder[iIndex]) { break; } } if(iIndex>0) { return FindFirst(pre+1,inorder,iIndex); } else { return FindFirst(pre+iIndex+1,inorder+iIndex+1,n-iIndex-1); } } int main() { #ifdef ONLINE_JUDGE #else freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int n; scanf("%d",&n); int pre[n],inorder[n]; for(int i=0;i<n;++i) scanf("%d",&pre[i]); for(int i=0;i<n;++i) scanf("%d",&inorder[i]); printf("%d ",FindFirst(pre,inorder,n)); return 0; }
1139题：有点难度，注意两点：1）0000是男生，-0000是女生。2）使用递归dfs最后一个case会超时，老老实实写几层循环去遍历吧！
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; #define N (10010) int cmp(const vector<int>& v1,const vector<int>& v2) { if(v1[0] != v2[0]) return v1[0] < v2[0]; return v1[1] < v2[1]; } vector<int> vssame[N]; //相同 vector<int> vsdif[N]; //不同 bool bVis[N]; int thepeo1,thepeo2; bool bSame; vector<vector<int> > res; void dfs() { res.clear(); memset(bVis,0,sizeof(bVis)); bVis[thepeo1] = true; for(int i=0;i<vssame[thepeo1].size();++i) { int id1 = vssame[thepeo1][i]; //printf("1:%04d ",id1); if(!bVis[id1]&&id1!=thepeo2) { bVis[id1] = true; vector<int> *pvs; if(bSame) pvs = &(vssame[id1]); else pvs = &(vsdif[id1]); for(int j=0;j<pvs->size();++j) { int id2 = (*pvs)[j]; //printf(" 2:%04d ",id2); if(!bVis[id2]&&id2!=thepeo2) { bVis[id2] = true; for(int k=0;k<vssame[id2].size();++k) { int id3 = vssame[id2][k]; //printf(" 3:%04d ",id3); if(id3==thepeo2) { vector<int> vi; vi.push_back(id1); vi.push_back(id2); res.push_back(vi); break; } } bVis[id2] = false; } } bVis[id1] = false; } } } int main() { #ifdef ONLINE_JUDGE #else freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int n,m; scanf("%d%d",&n,&m); for(int i=0;i<m;++i) { char str1[20],str2[20]; scanf("%s%s",str1,str2); int id1 = abs(atoi(str1)),id2=abs(atoi(str2)); if(str1[0]=='-'&&str2[0]=='-' ||str1[0]!='-'&&str2[0]!='-') { vssame[id1].push_back(id2); vssame[id2].push_back(id1); } else { vsdif[id1].push_back(id2); vsdif[id2].push_back(id1); } } int t; scanf("%d",&t); while(t--) { char str1[20],str2[20]; scanf("%s%s",str1,str2); if(str1[0]=='-'&&str2[0]=='-'||str1[0]!='-'&&str2[0]!='-') { bSame = true; } else { bSame = false; } thepeo1 = abs(atoi(str1)); thepeo2 = abs(atoi(str2)); dfs(); sort(res.begin(),res.end(),cmp); printf("%d ",res.size()); for(int i=0;i<res.size();++i) { vector<int>& vi = res[i]; for(int j=0;j<vi.size();++j) { if(j) printf(" "); printf("%04d",vi[j]); } printf(" "); } } return 0; }
1140题：题意很难理解。。。一句话后面数字是前面数字的描述。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<vector> #include<map> #include<set> #include<string> #include<algorithm> using namespace std; string getstr(string str) { string res=""; int len = str.length(); char c= str[0],cnt=0; for(int i=0;i<len;++i) { if(c==str[i]) ++cnt; else if(c!=str[i]) { res += c; char strcnt[1000]; sprintf(strcnt,"%d",cnt); res +=strcnt; c = str[i]; cnt = 1; } if(i==len-1) { res += c; char strcnt[1000]; sprintf(strcnt,"%d",cnt); res +=strcnt; } } return res; } //题意：后面的数字是对前面数字的描述 int main() { #ifdef ONLINE_JUDGE #else freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE char a[100]; int n; scanf("%s%d",&a,&n); string str = a; for(int i=1;i<n;++i) str=getstr(str); printf("%s ",str.c_str()); return 0; }
1142题：要求最大团：团是一组顶点，每个顶点两两可以组成边，最大团就是没有另外一个顶点与团里的点都能组成边。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; #define N (250) int Map[N][N]; int nv,ne; int main() { #ifdef ONLINE_JUDGE #else freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE scanf("%d%d",&nv,&ne); for(int i=0;i<ne;++i) { int a,b; scanf("%d%d",&a,&b); Map[a][b] = Map[b][a] = 1; } int t; scanf("%d",&t); while(t--) { int k; scanf("%d",&k); int a[k]; for(int i=0;i<k;++i) scanf("%d",&a[i]); bool bClique = true; for(int i=0;i<k;++i) for(int j=0;j<k;++j) { //要求每条边两两相交 if(i!=j&&Map[a[i]][a[j]]==0) { bClique = false; break; } } if(bClique) { bool bMax = true; for(int i=1;i<=nv;++i) { int j; for(j=0;j<k;++j) { if(Map[i][a[j]]==0) break; } if(j==k) { bMax = false; break; } } if(bMax) printf("Yes "); else printf("Not Maximal "); } else printf("Not a Clique "); } return 0; }
1143题：用map存储出现过的点，用于判断是否点存在。通过遍历先序，第一个出现满足a,b在val左右两边的就是他们的父节点。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; int main() { #ifdef ONLINE_JUDGE #else freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE map<int,bool> Map; int m,n; scanf("%d%d",&m,&n); int pre[n]; for(int i=0;i<n;++i) { scanf("%d",&pre[i]); Map[pre[i]] = true; } while(m--) { int a,b; scanf("%d%d",&a,&b); bool aexits = Map[a],bexits = Map[b]; if(!aexits&&!bexits) printf("ERROR: %d and %d are not found. ",a,b); else if(aexits&&bexits) for(int i=0;i<n;++i) { int val = pre[i]; if(val>a&&val<b || val<a&&val>b) { printf("LCA of %d and %d is %d. ",a,b,val); break; } else if(val==a|| val==b) { printf("%d is an ancestor of %d. ",val==a?a:b,val==a?b:a); break; } } else printf("ERROR: %d is not found. ",!aexits?a:b); } return 0; }
1144题：简单题，注意全部小于0是，应该输出1
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; int main() { #ifndef ONLINE_JUDGE freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int n; scanf("%d",&n); int a[n]; for(int i=0;i<n;++i) scanf("%d",&a[i]); sort(a,a+n); for(int i=1,j=0;j<n;++i) for(;j<n;++j) { if(a[j]==i) break; else if(a[j]>i) { printf("%d ",i); return 0; } } printf("%d ",max(a[n-1]+1,1)); //printf("%d ",a[n-1]); return 0; }
1145题：hash的二次探测，题目比较坑 ，算查找的总次数，一般二次探测0到size-1次没找到就好了，但是它而要多加一，所以循环要写到0到size；
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; bool isPrime(int n) { for(int i=2;i*i<=n;++i) if(n%i==0) return false; return true; } int getsize(int n) { while(!isPrime(n)) ++n; return n; } int main() { #ifndef ONLINE_JUDGE freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int msize,n,m; scanf("%d%d%d",&msize,&n,&m); msize = getsize(msize); int Hash[msize]; memset(Hash,0,sizeof(Hash)); for(int i=0;i<n;++i) { int a; scanf("%d",&a); int j; for(j=0;j<msize;++j) if(Hash[(a+j*j)%msize]==0) { Hash[(a+j*j)%msize] = a; break; } if(j>=msize) printf("%d cannot be inserted. ",a); } int t = m,cnt=0; while(t--) { int a; scanf("%d",&a); for(int i=0;i<=msize;++i) { if(++cnt&&(Hash[(a+i*i)%msize]==a||(Hash[(a+i*i)%msize]==0))) break; } } printf("%.1lf ",1.0*cnt/m); return 0; }
1146题：拓扑排序，了解拓扑排序的原理就行了。学习拓扑排序必须知道什么是入度，有向图中有边：顶点a到顶点b，则b的入度是1。所以拓扑排序是每次输出入度为0的顶点，并且把该顶点相连的边的另一个顶点的入度减一。
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; #define N (1100) int Map[N][N]; int indegree[N]; int n,m; bool isTopuSort(int* a) { int temp_indegree[N]; memcpy(temp_indegree,indegree,sizeof(temp_indegree)); for(int i=0;i<n;++i) { if(temp_indegree[a[i]]!=0) return false; for(int j=1;j<=n;++j) if(Map[a[i]][j]) --temp_indegree[j]; } return true; } int main() { #ifndef ONLINE_JUDGE freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE scanf("%d%d",&n,&m); for(int i=0;i<m;++i) { int a,b; scanf("%d%d",&a,&b); ++indegree[b]; Map[a][b] = 1; } int T; scanf("%d",&T); bool bFirst = true; for(int t=0;t<T;++t) { int a[n]; for(int i=0;i<n;++i) { scanf("%d",&a[i]); } if(isTopuSort(a)==false) { if(bFirst) bFirst= false; else printf(" "); printf("%d",t); } } printf(" "); return 0; }
1147题：简单题，可以用不构造树的方式做。（当然构树也行，只是会麻烦一点~） heap tree。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; #define LEFT(i) (2*i+1) #define RIGHT(i) (2*i+2) bool isMaxHeap(int* a,int curI,int n) { if(curI>=n) return true; int il = LEFT(curI),ir = RIGHT(curI); if(il<n&&a[curI]<a[il]) return false; if(ir<n&&a[curI]<a[ir]) return false; return isMaxHeap(a,il,n)&&isMaxHeap(a,ir,n); } bool isMinHeap(int* a,int curI,int n) { if(curI>=n) return true; int il = LEFT(curI),ir = RIGHT(curI); if(il<n&&a[curI]>a[il]) return false; if(ir<n&&a[curI]>a[ir]) return false; return isMinHeap(a,il,n)&&isMinHeap(a,ir,n); } bool bFirst; void post_travel(int* a,int curI,int n) { if(curI>=n) return; post_travel(a,LEFT(curI),n); post_travel(a,RIGHT(curI),n); if(bFirst) bFirst = false; else printf(" "); printf("%d",a[curI]); } int main() { #ifndef ONLINE_JUDGE freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int t,n; scanf("%d%d",&t,&n); while(t--) { int a[n]; for(int i=0;i<n;++i) scanf("%d",&a[i]); if(isMaxHeap(a,0,n)) printf("Max Heap "); else if(isMinHeap(a,0,n)) printf("Min Heap "); else printf("Not Heap "); bFirst = true; post_travel(a,0,n); printf(" "); } return 0; }
1148题：题目不好理解，有两个狼人，所有人中只有一狼一人说谎，输出两个狼人。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; int main(){ #ifndef ONLINE_JUDGE freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int n; scanf("%d",&n); int a[n+1]; for(int i=1;i<=n;++i) scanf("%d",&a[i]); for(int i=1;i<=n;++i){ for(int j=i+1;j<=n;++j){ int bwolf[n+1]; memset(bwolf,0,sizeof(bwolf)); bwolf[i] = bwolf[j] = 1; //i,j为狼 bool bOK = (a[i]>0)==bwolf[abs(a[i])]&&(a[j]<0)==bwolf[abs(a[j])]|| (a[i]<0)==bwolf[abs(a[i])]&&(a[j]>0)==bwolf[abs(a[j])]; for(int k=1,lie=0;k<=n&&bOK;++k){ if(i==k||j==k) continue; if((a[k]>0)==bwolf[abs(a[k])]){ ++lie; if(lie>1) bOK = false; } } if(bOK){ printf("%d %d ",i,j); return 0; } } } printf("No Solution "); return 0; }
1149题：简单题，两个物品放一起会爆炸，给出一串东西，判断这些东西有没有危险。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; set<int> misi[100010]; int main() { #ifndef ONLINE_JUDGE freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int n,m; scanf("%d%d",&n,&m); for(int i=0;i<n;++i) { int a,b; scanf("%d%d",&a,&b); misi[a].insert(b); misi[b].insert(a); } while(m--) { int k; scanf("%d",&k); vector<int> vi; bool bOK = true; for(int i=0;i<k;++i) { int a; scanf("%d",&a); if(bOK) { set<int>& sidag=misi[a]; for(int j=0;j<vi.size();++j) { if(sidag.find(vi[j])!=sidag.end()) { bOK = false; break; } } vi.push_back(a); } } printf("%s ",bOK?"Yes":"No"); } return 0; }
1150题：简单题
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; int main() { #ifndef ONLINE_JUDGE freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int n,m; scanf("%d%d",&n,&m); int Map[n+1][n+1]; memset(Map,0,sizeof(Map)); for(int i=0;i<m;++i) { int a,b,l; scanf("%d%d%d",&a,&b,&l); Map[a][b] = Map[b][a] = l; } int T; int minLen = 1<<30; int minPath = -1; scanf("%d",&T); for(int t=1;t<=T;++t) { printf("Path %d: ",t); int cn; scanf("%d",&cn); int istart,iend; bool bVis[n+1]; memset(bVis,0,sizeof(bVis)); bool bSimple = true; int cnt = 0,pre,len=0; int bNA = false; for(int i=0;i<cn;++i) { int a; scanf("%d",&a); if(i&&Map[pre][a]==0) bNA = true; else if(i) len += Map[pre][a]; if(i==0) istart = a; if(i==cn-1) iend = a; pre = a; if(bVis[a]==0) { bVis[a] = 1; ++cnt; } else if(i!=cn-1) bSimple = false; } if(bNA) printf("NA "); else printf("%d ",len); if(cnt!=n||bNA||istart!=iend) printf("(Not a TS cycle) "); else { if(minLen > len) { minLen = len; minPath = t; } printf("(%s) ",bSimple?"TS simple cycle":"TS cycle"); } } printf("Shortest Dist(%d) = %d ",minPath,minLen); return 0; }
1151题：用两个map，一个是把val转换成编号（从1开始，0表示没有），一个是把编号转换成val。中序输入的时候通过把val转换成编号，中序就是从小到大排序的，所以该树就变成搜索二叉树了。然后通过pre数组记录先序的编号，通过先前记录的map吧val转换成编号。所以该数时关于编号的搜索二叉树。然后接下来的做法就跟1143题一模一样了。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; int main() { #ifdef ONLINE_JUDGE #else freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int m,n; scanf("%d%d",&m,&n); map<int,int> valtoid; map<int,int> idtoval; int pre[n]; for(int i=0;i<n;++i) { int a; scanf("%d",&a); valtoid[a] = i+1; idtoval[i+1] = a; } for(int i=0;i<n;++i) { int a; scanf("%d",&a); pre[i] = valtoid[a]; } while(m--) { int a,b; scanf("%d%d",&a,&b); int aexits=valtoid[a],bexits=valtoid[b]; if(!aexits&&!bexits) printf("ERROR: %d and %d are not found. ",a,b); else if(!aexits||!bexits) printf("ERROR: %d is not found. ",!aexits?a:b); else { for(int i=0;i<n;++i) { int ia = valtoid[a],ib = valtoid[b]; if(pre[i]<ia&&pre[i]>ib||pre[i]<ib&&pre[i]>ia) { printf("LCA of %d and %d is %d. ",a,b,idtoval[pre[i]]); break; } else if(pre[i]==ia||pre[i]==ib) { printf("%d is an ancestor of %d. ",pre[i]==ia?a:b,pre[i]==ia?b:a); break; } } } } return 0; }
1152题：简单题，题目的tip也很详细。
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; #define L (1010) bool isPrime(long long int a) { if(a==0) return false; for(long long int i=2;i*i<=a&&i*i>0;++i) if(a%i==0) return false; return true; } int main() { #ifndef ONLINE_JUDGE freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int l,k; scanf("%d%d",&l,&k); char str[L]; scanf("%s",&str); string res=""; for(int i=0;i<l;++i) { res += str[i]; if(i+1>=k) { if(res.length()>k) res = res.substr(1,k); long long int a = atoll(res.c_str()); if(isPrime(a)) { printf("%s ",res.c_str()); return 0; } } } printf("404 "); return 0; }
1153题：简单题，不要每次搜索1类型的时候都去排序，不然会超时。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; struct PAT { string str; int sorce; string type; //考试类型 string sitnum; string date; string testeenum; void init(string str,int sorce) { this->str = str; this->sorce =sorce; int i = 0; sitnum = date = testeenum = ""; type = str[i++]; for(;i<4;++i) sitnum += str[i]; for(;i<10;++i) date += str[i]; for(;i<13;++i) testeenum+=str[i]; } }; int sitnum[1000]; PAT pats[10010]; int sitcmp(const int& a,const int& b) { if(sitnum[a]!=sitnum[b]) return sitnum[a]>sitnum[b]; return a<b; } int cmp(const int& a,const int& b) { PAT p1 = pats[a],p2=pats[b]; if(p1.sorce!=p2.sorce) return p1.sorce>p2.sorce; return strcmp(p1.str.c_str(),p2.str.c_str())<0; } int main() { #ifndef ONLINE_JUDGE freopen("test.txt","r",stdin); #endif // ONLINE_JUDGE int n,m; scanf("%d%d",&n,&m); map<string,vector<int> > searchmap[3]; for(int i=0;i<n;++i) { char str[30]; int sorce; scanf("%s%d",str,&sorce); pats[i].init(str,sorce); searchmap[0][pats[i].type].push_back(i); searchmap[1][pats[i].sitnum].push_back(i); searchmap[2][pats[i].date].push_back(i); } set<string> bsort; //关于1对应key的序列是否排序好了 for(int i=0;i<m;++i) { int opt; char str[30]; scanf("%d%s",&opt,str); printf("Case %d: %d %s ",i+1,opt,str); vector<int> &vi = searchmap[opt-1][str]; memset(sitnum,0,sizeof(sitnum)); if(vi.size()) { if(opt==1&&bsort.find(str)==bsort.end()) sort(vi.begin(),vi.end(),cmp),bsort.insert(str); int tsorce = 0; for(int j=0;j<vi.size();++j) { int index = vi[j]; if(opt==1) printf("%s %d ",pats[index].str.c_str(),pats[index].sorce); else if(opt==2) tsorce += pats[index].sorce; else if(opt==3) ++sitnum[atoi(pats[index].sitnum.c_str())]; } if(opt ==2) printf("%d %d ",vi.size(),tsorce); else if(opt==3) { vector<int> sitvc; for(int j=0;j<1000;++j) if(sitnum[j]) sitvc.push_back(j); sort(sitvc.begin(),sitvc.end(),sitcmp); for(int j=0;j<sitvc.size();++j) printf("%03d %d ",sitvc[j],sitnum[sitvc[j]]); } } else printf("NA "); } return 0; }
1154题：简单题
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; int main() { #ifndef ONLINE_JUDGE freopen("test.txt","r",stdin); #endif int n,m; scanf("%d%d",&n,&m); vector<int> Map[n]; for(int i=0;i<m;++i) { int a,b; scanf("%d%d",&a,&b); Map[a].push_back(b); } int t; scanf("%d",&t); while(t--) { set<int> si; int color[n]; for(int i=0;i<n;++i) { scanf("%d",&color[i]); si.insert(color[i]); } bool bOK = true; for(int i=0;i<n&&bOK;++i) for(int j=0;j<Map[i].size();++j) if(color[i]==color[Map[i][j]]) { bOK = false; break; } if(bOK) printf("%d-coloring ",si.size()); else printf("No "); } return 0; }
1155题：简单题，跟1147一样是heap tree。估计题目有点想出堆排序，但是又怕没人会堆排序所以搞成这样吧。。
#include<iostream> #include<cstdio> #include<set> #include<map> #include<vector> #include<iterator> #include<algorithm> #include<cstring> using namespace std; #define LEFT(i) (2*i+1) #define RIGHT(i) (2*i+2) struct Node { int val; Node* left; Node* right; Node(){left=right=NULL;} }; Node* Build(int *a,int i,int n) { if(i >= n) return NULL; Node* node = new Node(); node->val = a[i]; node->left = Build(a,LEFT(i),n); node->right = Build(a,RIGHT(i),n); return node; } void travel(Node* node,vector<Node*> vn) { vn.push_back(node); if(!node->left&&!node->right) { for(int i=0;i<vn.size();++i) { if(i) printf(" "); printf("%d",vn[i]->val); } printf(" "); } if(node->right) travel(node->right,vn); if(node->left) travel(node->left,vn); } bool bMaxHeap(int* a,int i,int n) { if(i>=n) return true; if(LEFT(i)<n&&a[i]<a[LEFT(i)]) return false; if(RIGHT(i)<n&&a[i]<a[RIGHT(i)]) return false; return bMaxHeap(a,LEFT(i),n)&&bMaxHeap(a,RIGHT(i),n); } bool bMinHeap(int* a,int i,int n) { if(i>=n) return true; if(LEFT(i)<n&&a[i]>a[LEFT(i)]) return false; if(RIGHT(i)<n&&a[i]>a[RIGHT(i)]) return false; return bMinHeap(a,LEFT(i),n)&&bMinHeap(a,RIGHT(i),n); } int main() { #ifndef ONLINE_JUDGE freopen("test.txt","r",stdin); #endif int n; scanf("%d",&n); int a[n]; for(int i=0;i<n;++i) scanf("%d",&a[i]); Node* root = Build(a,0,n); vector<Node*> vn; travel(root,vn); if(bMaxHeap(a,0,n)) printf("Max Heap "); else if(bMinHeap(a,0,n)) printf("Min Heap "); else printf("Not Heap "); return 0; }