PAT 1046-1057
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1046. Shortest Distance(20)
原题地址：http://www.patest.cn/contests/pat-a-practise/1046
解题思路：无他。求距离的题目，最好选一个参考点保存各节点的位置，这样两节点间的距离由位置差即可求得。
代码如下：
#include <cstdio>#include <cstring>#define MAXN 100005int dist[MAXN];int TD;int N, M;int main(){// freopen("1046.txt","r",stdin); TD = 0; memset(dist,0,sizeof(dist)); scanf("%d",&N); for(int i = 0; i < N; i++) {scanf("%d",dist+i+1); TD += dist[i+1]; dist[i+1] = TD;} scanf("%d",&M); for(int i = 0; i < M; i++){ int f, t; scanf("%d%d",&f,&t); if(f > t) { int tm = t; t = f; f = tm;}// int dm = 0; if(dist[t-1] - dist[f-1] <= TD + dist[f-1] - dist[t-1]) printf("%d\n",dist[t-1] - dist[f-1]); else printf("%d\n",TD + dist[f-1] - dist[t-1]);// printf("%d,%d,%d\n",TD,dist[f-1],dist[t-1]); } return 0;}1047. Student List for Course (25)
原题地址：http://www.patest.cn/contests/pat-a-practise/1047
解题思路：用vector保存每门课的学生名单即可。
代码如下：
#include <cstdio>#include <cstring>#include <string>#include <vector>#include <algorithm>#define MAXK 2505using namespace std;vector<string> co[MAXK];int N,K;int main(){// freopen("1047.txt","r",stdin); scanf("%d%d",&N,&K); for(int i = 0; i < N; i++){ char name[8]; int c; scanf("%s%d",name,&c); string n(name); for(int i = 0; i < c; i++){ int t; scanf("%d",&t); co[t].push_back(n); } } for(int i = 1; i <= K; i++){ int m = co[i].size(); printf("%d %d\n",i,m); if(m != 0){ sort(co[i].begin(),co[i].end()); for(int j = 0; j < m; j++) printf("%s\n", co[i][j].c_str()); } }}1048. Find Coins (25)
解题思路：开数组记录不同面值银币的个数，再从头往后找符合的组合即可。
原题地址：http://www.patest.cn/contests/pat-a-practise/1048
代码如下：
#include <cstdio>#include <cstring>#define MAXN 1010int cou[MAXN];int N, M;int main(){// freopen("1048.txt","r",stdin); memset(cou,0,sizeof(cou)); scanf("%d%d",&N,&M); for(int i = 0 ; i < N; i++){ int co; scanf("%d",&co); cou[co]++; } for(int i = 1 ; i <= 500 ; i++){ if( i*2 < M && cou[i] > 0 && cou[M-i] >0){ printf("%d %d",i,M-i); return 0; }else if( i*2 == M && cou[i] > 1){ printf("%d %d", i, i); return 0; } } printf("No Solution"); return 0;}1049. Counting Ones (30)
原题地址：http://www.patest.cn/contests/pat-a-practise/1049
解题思路：我的思路比较蠢，将数字切分成整十整百整千部分考虑，比较繁琐。
注：（网上有好的方法，即考虑每一位上1出现的个数，找出规律，过程简洁明了。具体可参考：http://blog.csdn.net/tiantangrenjian/article/details/19908885）
代码如下：
#include <cstdio>#include <cmath>int N;int cou;int generate(int m){ int co = 0; for(int i = 0; i < m; i++){ co = pow(10,i) + 10*co; }// printf("%d,return gene: %d\n",m,co); return co;}int count(int n){// printf("n:%d\n",n); int m = 0; int tmp = n; int co = 0; for(; m <= 9 ;m++){ if(tmp/10 == 0) break; tmp /= 10; } // printf("tmp:%d\n",tmp); if(tmp > 1) co += ( pow(10,m) + count(n - tmp*pow(10,m)) + (tmp) * generate(m)); else if(tmp == 1) co += (n - tmp*pow(10,m) + 1 + count(n - tmp*pow(10,m)) + generate(m)); // printf("count %d: %d\n", n, co); return co;}int main(){ scanf("%d",&N); cou = count(N); printf("%d\n",cou); return 0;}1050. String Subtraction (20)
原题地址：http://www.patest.cn/contests/pat-a-practise/1050
解题思路：记录s2中每个字符出现与否。我的坑点在于，一开始将strlen()函数写进循环判断，导致超时。
代码如下：
#include <cstdio>#include <cstring>#define MAXN 10010char data[MAXN];char ab[MAXN];int flag[100];int main(){// freopen("1050.txt","r",stdin); gets(data); memset(flag,0,sizeof(flag)); gets(ab);// int cou = 0; int l = strlen(ab); for(int i = 0; i < l; i++){ int id = ab[i] - ' ';// if(flag[id] == 0) cou++; flag[id] = 1; } // printf("%s",data); l = strlen(data); for(int i = 0; i < l; i++){ char c = data[i]; if(!flag[c-' ']) printf("%c",c); } return 0;}1051. Pop Sequence (25)
原题地址：http://www.patest.cn/contests/pat-a-practise/1051
解题思路：用stack保存过渡栈内容。。
代码如下：
#include <cstdio>#include <algorithm>#include <stack>#define MAXN 1005using namespace std;stack<int> st;int sq[MAXN];int M, N, K;void clear(){ while(!st.empty()) st.pop(); }int isPop(){ int co = 0; clear(); for(int i = 0; i < N; i++){ if(st.size() < M && i+1 == sq[co]){ co++;// printf("sq[%d]:%d\n",co,sq[co]); // if(!st.empty()) printf("st:%d\n",st.top()); while(!st.empty() && st.top() == sq[co]){ co++; st.pop(); } }else if(st.size() < M) st.push(i+1); else return 0; } if(st.empty()) return 1; else return 0; }int main(){// freopen("1051.txt","r",stdin); scanf("%d%d%d",&M,&N,&K); for(int i = 0; i < K; i++){ for(int j = 0; j < N; j++){ scanf("%d",sq+j); } if(isPop()) printf("YES\n"); else printf("NO\n"); } }1052. Linked List Sorting (25)
原题地址：http://www.patest.cn/contests/pat-a-practise/1052解题思路：读入链表元素，进行排序。注意节点的指向下一地址。
代码如下：
#include <cstdio>#include <cstring>#include <vector>#include <algorithm>#define MAXN 100005using namespace std;typedef struct{ int add; int val; int nest;}node;vector<node> vn;node n[MAXN];int val[MAXN];int nest[MAXN];int N, sadd;int cmp(const node n1, const node n2){ return n1.val < n2.val;}int main(){// freopen("1052.txt","r",stdin); memset(nest,-1,sizeof(nest)); scanf("%d%d",&N,&sadd); for(int i = 0; i < N; i++){ int s,v,t; scanf("%d%d%d",&s,&v,&t); nest[s] = t; val[s] = v; } int s = sadd; int cou = 0; while(s!=-1){ n[s].add = s; n[s].val = val[s]; n[s].nest = nest[s]; vn.push_back(n[s]); s = nest[s]; cou++; } sort(vn.begin(),vn.end(),cmp); if(cou > 0) printf("%d %.5d\n", cou,vn[0].add); else printf("0 -1"); for(int i = 0; i < cou-1; i++){ printf("%.5d %d %.5d\n", vn[i].add, vn[i].val, vn[i+1].add); } if(cou > 0) printf("%.5d %d %d\n", vn[cou-1].add, vn[cou-1].val, -1); return 0;}1053. Path of Equal Weight (30)
原题地址：http://www.patest.cn/contests/pat-a-practise/1053解题思路：dfs+回溯即可。
代码如下：
#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;#define MAXN 105int chd[MAXN][MAXN];int val[MAXN];int isFa[MAXN];vector<vector<int> > route;vector<int> ro;int N, M, S;int curm;int cmp(const vector<int> v1, const vector<int> v2){ int m = v1.size(); if(m > v2.size()) m = v2.size(); for(int i = 0; i < m; i++){ if(v1[i] == v2[i]) continue; else return v1[i] > v2[i]; } return 0;}int find(int root){ ro.push_back(val[root]); curm += val[root]; if(curm == S && !isFa[root]){ vector<int> v; v.assign(ro.begin(),ro.end()); route.push_back(v); }else if(curm < S){ for(int i = 0; i < N; i++){ if(chd[root][i]) find(i); } } ro.pop_back(); curm -= val[root]; return 0;}int main(){// freopen("1053.txt","r",stdin); memset(chd,0,sizeof(chd)); memset(val,0,sizeof(val)); memset(isFa,0,sizeof(isFa)); scanf("%d%d%d",&N, &M, &S); for(int i = 0; i < N; i++) scanf("%d",val+i); for(int i = 0; i < M; i++){ int id,num; scanf("%d%d",&id,&num); isFa[id] = 1; for(int j = 0; j < num; j++){ int c; scanf("%d",&c); chd[id][c] = 1; } } int root = 0; curm = 0; find(root); sort(route.begin(),route.end(),cmp); for(int i = 0; i < route.size(); i++){ vector<int> v = route[i]; int flag = 1; for(int j = 0; j < v.size(); j++){ if(flag) flag = 0; else printf(" "); printf("%d", v[j]); } printf("\n"); }}1054. The Dominant Color (20)
原题地址：http://www.patest.cn/contests/pat-a-practise/1054
解题思路：直接遍历找出现最多的元素也能ac。
（注：说一下巧妙的解法： 首先用num 和 count 分别来保存出现最多的数和次数。每次读入一个数，若 count 等于0，则将num赋值读入的数，count ++； 否则，若读入的数与num相等，count++，若读入的数与num 不同，则count--； 这样最后保存的额num一定是出现次数大于一半的那个数。）
代码如下：
#include <cstdio>#include <cstring>#include <algorithm>#define MAXN 805*605using namespace std;int data[MAXN];int M,N;int main(){// freopen("1054.txt","r",stdin); scanf("%d%d",&M,&N); for(int i = 0; i < M*N; i++) scanf("%d",data+i); sort(data,data+M*N); printf("%d\n",data[M*N/2]);}1055. The World's Richest (25)
原题地址：http://www.patest.cn/contests/pat-a-practise/1055
解题思路：用stl中的sort即可。
代码如下：
#include <cstdio>#include <algorithm>#include <cstring>#define MAXN 100005using namespace std;typedef struct{ char name[10]; int age; int worth;}person;person p[MAXN];int N,K;int cmpN(const char* a, const char *b){ int lena = strlen(a); int lenb = strlen(b); for(int i = 0; i < lena && i < lenb; i++){ if(a[i] == b[i]) continue; else return a[i] < b[i]; } return lena > lenb;}int cmp(const person p1, const person p2){ if(p1.worth == p2.worth){ if(p1.age == p2.age) return cmpN(p1.name, p2.name); else return p1.age < p2.age; } else return p1.worth > p2.worth;}int main(){// freopen("1055.txt","r",stdin); scanf("%d%d",&N,&K); for(int i = 0; i < N; i++){ scanf("%s%d%d",p[i].name,&p[i].age,&p[i].worth); } sort(p,p+N,cmp); for(int i = 0; i < K; i++){ printf("Case #%d:\n", i+1); int m,f,t; scanf("%d%d%d",&m,&f,&t); int flag = 1; int index = 0; while(m>0 && index < N){ if(p[index].age >= f && p[index].age <= t){ m--; printf("%s %d %d\n", p[index].name, p[index].age, p[index].worth); flag = 0; } index++; } if(flag) printf("None\n"); }}1056. Mice and Rice (25)
原题地址：http://www.patest.cn/contests/pat-a-practise/1056
解题思路：下一回合晋级人数+1即为当前排名。
代码如下：
#include <cstdio>#include <queue>#include <cstring>#define MAXN 1005using namespace std;queue<int> q1;queue<int> q2;int data[MAXN];int order[MAXN];int rak[MAXN];int Np,Ng;int main(){ // freopen("1056.txt","r",stdin); scanf("%d%d",&Np,&Ng); memset(rak,0,sizeof(rak)); for(int i = 0; i< Np; i++) scanf("%d",data+i); for(int i = 0; i< Np; i++) {scanf("%d",order+i); q1.push(order[i]);} int flag = 1; // printf("%d:%d\n",Np,Ng); while(!q1.empty() || !q2.empty()){ if(flag){ int a[Ng]; int mx = -1; int indx = -1; int co = 0; int si = q1.size(); int rk = si/Ng + (si%Ng > 0) + 1; if(si == 1){ rak[q1.front()] = 1; break; } for(int i = 0; i < si; i++){ co++; a[co-1] = q1.front(); rak[a[co-1]] = rk; q1.pop(); if(mx < data[a[co-1]]) { mx = data[a[co-1]]; indx = a[co-1];} if( co == Ng ){ co = 0; q2.push(indx); indx = -1; mx = -1; } } for(int i = 0; i < co; i++){ rak[a[i]] = rk; if(mx < data[a[i]]) { mx = data[a[i]]; indx = a[i];} } if(indx > -1) q2.push(indx); flag = 0; }else{ int a[Ng]; int mx = -1; int indx = -1; int co = 0; int si = q2.size(); int rk = si/Ng + (si%Ng > 0) + 1; if(si == 1){ rak[q2.front()] = 1; break; } for(int i = 0; i < si; i++){ co++; a[co-1] = q2.front(); rak[a[co-1]] = rk; q2.pop(); if(mx < data[a[co-1]]) { mx = data[a[co-1]]; indx = a[co-1];} if( co == Ng ){ co = 0; q1.push(indx); indx = -1; mx = -1; } } for(int i = 0; i < co; i++){ if(mx < data[a[i]]) { mx = data[a[i]]; indx = a[i];} rak[a[i]] = rk; } if(indx > -1) q1.push(indx); flag = 1; } } int fg = 1; for(int i = 0; i < Np; i++){ if(fg) fg = 0; else printf(" "); printf("%d", rak[i]); } return 0;}1057. Stack (30)
原题地址：http://www.patest.cn/contests/pat-a-practise/1057
解题思路：用树状数组来找中值。 （加上二分查找会更快。）
代码如下：
#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#define MAXN 100005using namespace std;int st[MAXN];int mdt[MAXN];int sz;int N;int getSum(int i){ int s = 0; while(i > 0){ s += mdt[i]; i -= i&(-i); } return s; }int query(int l, int h){ if(l == h) return l; int m = (l+h)/2; if(getSum(m) < (sz+1)/2) return query(m+1,h); else return query(l,m);}void update(int i, int value){ while(i < MAXN){ mdt[i] += value; i += i&(-i); } }void doPop(){ if(sz == 0) printf("Invalid\n"); else{ printf("%d\n",st[sz-1]); update(st[sz-1],-1); sz--; }}void doPeek(){ if(sz == 0) printf("Invalid\n"); else{ printf("%d\n",query(1,MAXN-1)); }}void doPush(int n){ st[sz] = n; update(n,1); sz++;} int main(){// freopen("1057.txt","r",stdin); sz = 0; memset(mdt,0,sizeof(mdt)); scanf("%d",&N); for(int i = 0; i < N; i++){ char s[15]; scanf("%s",s); if(s[1] == 'o') doPop(); else if(s[1] == 'e') doPeek(); else{ int n; scanf("%d",&n); doPush(n); } } return 0 } 0 0
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