Test20171009 考试总结 NOIP模拟赛

题目难度合适，区分度适中，但是本人水平不佳，没有拿到满意的分数。

T1(matrix)

一种比较容易想到的想法是枚举起点求出最长全1串做预处理，这是O(n^2)的。

接着枚举列起点，列终点，通过后缀和维护矩形的高。这也是O(n^2)的。

实现的细节看代码：

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>

 #define max(a,b) (((a)>(b))?(a):(b))

 const int maxn = ;

 int suf[maxn];//houzhuihe
 int mat[maxn][maxn];//matrix
 int len[maxn][maxn];//from some number ,the longest number

 int n,m;

 void read(){
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;i++)
     for(int j=;j<=m;j++)
         scanf("%1d",&mat[i][j]);
     for(int i=;i<=n;i++)
     for(int j=m;j>=;j--)
         len[i][j] = (mat[i][j]==?len[i][j+]+:);
 }

 void work(){
     int ans = ;
     for(int i=;i<=m;i++){
     for(int j=;j<=;j++)suf[j]=;
     for(int j=;j<=n;j++) suf[len[j][i]]++;
     for(int j=;j>=;j--) suf[j] = suf[j+]+suf[j];
     for(int j=;j<=;j++) ans = max(ans,j*suf[j]);
     }
     printf("%d",ans);
 }

 int main(){
     freopen("matrix.in","r",stdin);
     freopen("matrix.out","w",stdout);
     read();work();return ;
 }

T2(present)

　　这道题是一道完全背包恰好塞满的判断。对于希望达到的w，取单个价值最低的p。令剩下的构成费用x，x+kp=w.取最小的x即可。建图跑最短路。

 #include<iostream>
 #include<cstdio>
 #include<queue>
 #include<cstring>
 #include<cstdlib>
 #include<climits>
 #include<algorithm>
 using namespace std;

 struct edge{
     int to,w;
 };
 bool operator <(edge a,edge b){return a.w<b.w;}

 int n,m,num,mn;
 int p[],a[];
 int dist[];
 vector <edge> g[];

 void read(){
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;i++) scanf("%d",&p[i]);
     for(int j=;j<=m;j++) scanf("%d",&a[j]);
     sort(p+,p+n+);
     mn = p[];
 }

 int vis[];
 void spfa()
 {
     memset(dist,0x7f,sizeof(dist));
     queue<int> q;
     q.push();
     dist[]=;vis[]=;
     while(!q.empty()){
     int x=q.front();q.pop();
     vis[x]=;
     for(int i=;i<=n;i++){
         if(dist[(x+p[i])%mn]>dist[x]+p[i]){
         dist[(x+p[i])%mn]=dist[x]+p[i];
         if(!vis[(x+p[i])%mn]){
             q.push((x+p[i])%mn);
             vis[(x+p[i])%mn]=;
         }
         }
     }
     }
 }
 void work(){
     spfa();
     int ans=;
     for(int i=;i<=m;i++){
     if(dist[a[i]%mn]<=a[i])ans++;
     }
     printf("%d",ans);
 }

 int main(){
     freopen("present.in","r",stdin);
     freopen("present.out","w",stdout);
     read();
     work();
     return ;
 }

T3(mahjong)

　　搜索简单题，期望得分100实际得分0.。。

　　

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<algorithm>
 #pragma GCC optimize(2)
 using namespace std;

 int tb[],flag,what;
 int pd_double[]={,,,,,,,,,,,,,};
 int ans[],num;

 int str_num(char ch[]){
     switch(ch[]){
     case 'm':{return ch[]-'';}
     case 's':{return ch[]-''+;}
     case 'p':{return ch[]-''+;}
     case 'c':{return ch[]-''+;}
     }
 }
 void read(){
     memset(tb,,sizeof(tb));
     memset(ans,,sizeof(ans));
     for(int i=;i<=;i++){
     char ch[]; scanf("%s",ch);
     tb[str_num(ch)]++;
     }
 }

 int contry_no_double(){
     int etnum=;
     for(int i=;i<=;i++){
     if(!tb[pd_double[i]]){
         if(flag)return ; else{flag = ;what = pd_double[i];}
     }
     etnum+=tb[pd_double[i]];
     }
     if(flag == &&etnum==){ans[what] = ;return ;}
     else{
     printf("13 1m 9m 1s 9s 1p 9p 1c 2c 3c 4c 5c 6c 7c\n");
     return ;
     }
 }
 void seven_double(){
     int dnum = ,rec;
     for(int i=;i<=;i++){
     if(tb[i] > )return;
     if(tb[i] == )dnum++;
     if(tb[i] == )rec = i;
     }
     if(dnum == )ans[rec]=;
 }

 int dd[],nd;
 void dfs(int dep,int last,int deal){
     while(!tb[last]&&last<=)last++;
     if(deal == ){
     if(flag == )
         if(dd[]==dd[]&&nd==){
         ans[what]=;return;
         }
     if(flag==){
         if(nd==){ans[dd[]]=;return;}
         if(nd==){
         sort(dd+,dd+);
         if(dd[]==dd[]&&dd[]==dd[]){ans[dd[]]=;ans[dd[]]=;}
         if(dd[]-dd[]==&&dd[]==dd[]&&dd[]/!=){ans[dd[]-]=;ans[dd[]+]=;}
         if(dd[]==dd[]&&dd[]-dd[]==&&dd[]/!=){ans[dd[]+]=;ans[dd[]-]=;}
         }
         return;
     }
     }
     if(tb[last]&&tb[last+]&&tb[last+]&&last/!=){
     tb[last]--;tb[last+]--;tb[last+]--;
     dfs(dep+,last,deal+);
     tb[last]++;tb[last+]++;tb[last+]++;
     }
     if(tb[last]>=){
     tb[last]-=;
     dfs(dep+,last,deal+);
     tb[last]+=;
     }
     if(nd<){
     int cc[];for(int i=;i<=nd;i++)cc[i]=dd[i];
     cc[nd+]=last;
     sort(cc+,cc+nd+);
     int fff = ;
     if(nd<)fff=;
     if(!fff){
         tb[last]--;dd[++nd]=last;
         dfs(dep+,last,deal+);
         tb[last]++;nd--;
     }
     }
     if(tb[last]==&&(!flag)){
     tb[last]-=;flag = ;what = last;
     dfs(dep+,last+,deal+);
     flag = ;tb[last]+=;
     }
     if(last%==||last%==||last/==)return;
     if((bool)tb[last]+(bool)tb[last+]+(bool)tb[last+]== && (!flag)){
     if(!tb[last+]){
         tb[last]--;tb[last+]--;flag = ;what=last+;
         dfs(dep+,last,deal+);
         tb[last]++;tb[last+]++;flag = ;
     }
     if(!tb[last+]){
         tb[last]--;tb[last+]--;flag = ;what = last+;
         dfs(dep+,last,deal+);
         tb[last]++;tb[last+]++;flag = ;
     }
     }
 }

 void work(){
     flag = ;num=;
     int fg = contry_no_double();
     if(fg)return;
     flag = ;
     seven_double();
     dfs(,,);
     int a1=;
     for(int i=;i<=;i++){
     if(i%==)continue;
     if(ans[i]&&tb[i]==)ans[i]=;
     if(ans[i])a1++;
     }
     if(a1==){puts("Nooten");}
     else{
     printf("%d ",a1);
     for(int i=;i<=;i++) if(ans[i])printf("%dm ",i);
     for(int i=;i<=;i++) if(ans[+i])printf("%ds ",i);
     for(int i=;i<=;i++) if(ans[+i])printf("%dp ",i);
     for(int i=;i<=;i++) if(ans[+i])printf("%dc ",i);
     puts("");
     }
 }

 int main(){
     freopen("mahjong.in","r",stdin);
     freopen("mahjong.out","w",stdout);
     int t; scanf("%d",&t);
     while(t--){read();work();}
     return ;
 }