试题 A: 平方和
思路：
暴力模拟跑一遍即可，答案是2658417853
代码：
#include<bits/stdc++.h> using namespace std; int main() { long long cnt1=0,cnt2=0,cnt3=0; long long ans=0,i,j,t; for(i=1;i<=2019;i++) { j=i; int flag=0; while(j!=0) { if(j%10==0||j%10==1||j%10==2||j%10==9) { flag=1;break; } j/=10; } if(flag) { ans+=i*i; cout<<i<<endl; } } cout<<ans<<endl; return 0; }
试题 B: 数列求值
思路：
一个斐波那契数列的变种，因为项数很多而且求最后四位数组所以也不用想太多，答案是4659
代码：
#include<bits/stdc++.h> using namespace std; #define int long long int a[4]; signed main() { a[1]=1,a[2]=1,a[3]=1; int i,j,ans; for(i=4;i<=20190324;i++) { a[4]=(a[1]%10000+a[2]%10000+a[3]%10000)%10000; a[1]=a[2]%10000; a[2]=a[3]%10000; a[3]=a[4]%10000; } cout<<a[4]<<endl; return 0; }
试题 C: 最大降雨量
思路：
其实不难想到49个法术每天都要用也就是七个星期，而七个星期每次都取中位数，那么贪心的每次都想取最大，所以模拟一下，第一周取最小的1,2,3然后取个最大的49,48,47,46，再排序，取七次即可。最后答案34
#include<bits/stdc++.h> using namespace std; #define int long long signed main() { vector<int >ans[7]; int min1=1,max1=49,ans1=0; int i,j; for(i=0;i<7;i++) { while(ans[i].size()!=7) { if(ans[i].size()<3){ ans[i].push_back(min1++); } else { ans[i].push_back(max1--); } } sort(ans[i].begin(),ans[i].end()); for(j=0;j<ans[i].size();j++) cout<<ans[i][j]<<" "; cout<<endl; ans1+=ans[i][3]; cout<<ans[i][3]<<endl; } cout<<ans1<<endl; return 0; }
试题 D: 迷宫
思路：
发现是从起点到终点的最短路，所以首选bfs，然后要求路径，这里处理一下，把每个点是从哪个点过来的记录一下，然后输出的时候再从尾到头找回去，记得输出要翻转字符串。
代码：
#include<bits/stdc++.h> using namespace std; char mp[55][55]; int dx[5]= {0,1,0,0,-1},dy[5]= {0,0,-1,1,0}; bool vis[55][55]; int pre[55*55+1000]; int main() { string ly; int n,m,i,j,t; pair<int,int >PI; cin>>n>>m; getchar(); for(i=1; i<=n; i++) { for(j=1; j<=m; j++) { cin>>mp[i][j]; } } queue<pair<int,int >>qq; map<pair<int,int >,int >ans; map<pair<int,int >,pair<int,int >>edges; qq.push({1,1}); while(!qq.empty()) { pair<int,int > aa=qq.front(); qq.pop(); int x1=aa.first,y1=aa.second; if(x1==n&&y1==m) { cout<<ans[ {x1,y1}]<<endl; } for(i=1; i<=4; i++) { int x2=x1+dx[i],y2=y1+dy[i]; if(vis[x2][y2]==0&&mp[x2][y2]=='0'&&x2>=1&&x2<=n&&y1>=1&&y2<=m) { vis[x2][y2]=1; ans[ {x2,y2}]=ans[ {x1,y1}]+1; edges[ {x2,y2}].first=x1; edges[ {x2,y2}].second=y1; qq.push({x2,y2}); } } } int edx=n,edy=m; string ss=""; map<int ,char >qm; qm[1]='D',qm[2]='L',qm[3]='R',qm[4]='U'; while(1) { int x1=edges[ {edx,edy}].first; int y1=edges[ {edx,edy}].second; for(i=1;i<=4;i++) { if(x1+dx[i]==edx&&y1+dy[i]==edy) { ss+=qm[i]; break; } } edx=x1,edy=y1; if(edx==1&&edy==1){ break; } } reverse(ss.begin(),ss.end()); cout<<ss<<endl; return 0; }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
试题 E: RSA 解密
思路：
试题 F: 完全二叉树的权值
思路：
模拟找节点找深度就行，比较简单
#include<bits/stdc++.h> using namespace std; #define int long long signed main() { int N,i,j; vector<int>ans[555]; cin>>N; int cnt=0,sum1=1; for(i=1;i<=N;i++) { cin>>j; if(i<=sum1) { ans[cnt].push_back(j); } else { ++cnt; sum1+=(1<<cnt); ans[cnt].push_back(j); } } int max1=0,anss; for(i=0;i<=cnt;i++) { int ss1=0; for(j=0;j<ans[i].size();j++) { ss1+=ans[i][j]; } if(ss1>max1) { max1=ss1; anss=i+1; } } cout<<anss<<endl; return 0; }
试题 G: 外卖店优先级
思路：
一个模拟贪心题，观察到时间不是很长，所以模拟时间跑就行，不过这里细节挺多的，要注意出队入队的条件，读题要读清楚。
#include<bits/stdc++.h> using namespace std; const int maxn=1e5+1000; struct node { int pret,levl; bool f1; }mo[maxn]; struct node1 { int stt,idd; }qury[maxn]; bool cmp1(node1 a,node1 b ) { if(a.stt!=b.stt) return a.stt<b.stt; else return a.idd<b.idd; } int main() { int n,m,i,j,ts,id,t; cin>>n>>m>>t; for(i=0;i<m;i++) { cin>>qury[i].stt>>qury[i].idd; } sort(qury,qury+m,cmp1); for(i=0;i<m;i++) { ts=qury[i].stt,id=qury[i].idd; if(ts>mo[id-1].pret) mo[id-1].levl=max(mo[id-1].levl-(ts-mo[id-1].pret-1),0); if(mo[id-1].f1==1&&mo[id-1].levl<=3) mo[id-1].f1=0; mo[id-1].levl+=2; if(mo[id-1].f1==0&&mo[id-1].levl>5) mo[id-1].f1=1; mo[id-1].pret=ts; } int ans=0; for(i=0;i<maxn;i++) { mo[i].levl=max(0,mo[i].levl-(t-mo[i].pret)); if(mo[i].f1==0&&mo[i].levl>5) mo[i].f1=1; if(mo[i].f1==1&&mo[i].levl<=3) mo[i].f1=0; if(mo[i].f1==1) ans++; } cout<<ans<<endl; return 0; }
试题 H: 修改数组
思路：
真是没想到是并查集做法，待补
试题 I: 糖果
思路：
不会，二分贪心跑了几分
试题 J: 组合数问题
思路:
不会
