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 = 1024;

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=1;i<=n;i++)
    for(int j=1;j<=m;j++)
        scanf("%1d",&mat[i][j]);
    for(int i=1;i<=n;i++)
    for(int j=m;j>=1;j--)
        len[i][j] = (mat[i][j]==1?len[i][j+1]+1:0);
}

void work(){
    int ans = 0;
    for(int i=1;i<=m;i++){
    for(int j=0;j<=1000;j++)suf[j]=0;
    for(int j=1;j<=n;j++) suf[len[j][i]]++;
    for(int j=1000;j>=0;j--) suf[j] = suf[j+1]+suf[j];
    for(int j=1;j<=1000;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 0;
}

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[520],a[310000];
int dist[31000];
vector <edge> g[11000];

void read(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%d",&p[i]);
    for(int j=1;j<=m;j++) scanf("%d",&a[j]);
    sort(p+1,p+n+1);
    mn = p[1];
}

int vis[33000];
void spfa()
{
    memset(dist,0x7f,sizeof(dist));
    queue<int> q;
    q.push(0);
    dist[0]=0;vis[0]=1;
    while(!q.empty()){
    int x=q.front();q.pop();
    vis[x]=0;
    for(int i=1;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]=1;
        }
        }
    }
    }
}
void work(){
    spfa();
    int ans=0;
    for(int i=1;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 0;
}

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[60],flag,what;
int pd_double[15]={0,1,9,11,19,21,29,31,32,33,34,35,36,37};
int ans[60],num;

int str_num(char ch[]){
    switch(ch[1]){
    case 'm':{return ch[0]-'0';}
    case 's':{return ch[0]-'0'+10;}
    case 'p':{return ch[0]-'0'+20;}
    case 'c':{return ch[0]-'0'+30;}
    }
}
void read(){
    memset(tb,0,sizeof(tb));
    memset(ans,0,sizeof(ans));
    for(int i=1;i<=13;i++){
    char ch[4]; scanf("%s",ch);
    tb[str_num(ch)]++;
    }
}

int contry_no_double(){
    int etnum=0;
    for(int i=1;i<=13;i++){
    if(!tb[pd_double[i]]){
        if(flag)return 0; else{flag = 1;what = pd_double[i];}
    }
    etnum+=tb[pd_double[i]];
    }
    if(flag == 1&&etnum==13){ans[what] = 1;return 0;}
    else{
    printf("13 1m 9m 1s 9s 1p 9p 1c 2c 3c 4c 5c 6c 7c
");
    return 1;
    }
}
void seven_double(){
    int dnum = 0,rec;
    for(int i=1;i<=37;i++){
    if(tb[i] > 2)return;
    if(tb[i] == 2)dnum++;
    if(tb[i] == 1)rec = i;
    }
    if(dnum == 6)ans[rec]=1;
}

int dd[8],nd;
void dfs(int dep,int last,int deal){
    while(!tb[last]&&last<=37)last++;
    if(deal == 13){
    if(flag == 1)
        if(dd[2]==dd[1]&&nd==2){
        ans[what]=1;return;
        }
    if(flag==0){
        if(nd==1){ans[dd[1]]=1;return;}
        if(nd==4){
        sort(dd+1,dd+5);
        if(dd[1]==dd[2]&&dd[3]==dd[4]){ans[dd[1]]=1;ans[dd[3]]=1;}
        if(dd[2]-dd[1]==1&&dd[3]==dd[4]&&dd[2]/10!=3){ans[dd[1]-1]=1;ans[dd[2]+1]=1;}
        if(dd[1]==dd[2]&&dd[4]-dd[3]==1&&dd[4]/10!=3){ans[dd[4]+1]=1;ans[dd[3]-1]=1;}
        }
        return;
    }
    }
    if(tb[last]&&tb[last+1]&&tb[last+2]&&last/10!=3){
    tb[last]--;tb[last+1]--;tb[last+2]--;
    dfs(dep+1,last,deal+3);
    tb[last]++;tb[last+1]++;tb[last+2]++;
    }
    if(tb[last]>=3){
    tb[last]-=3;
    dfs(dep+1,last,deal+3);
    tb[last]+=3;
    }
    if(nd<4){
    int cc[8];for(int i=1;i<=nd;i++)cc[i]=dd[i];
    cc[nd+1]=last;
    sort(cc+1,cc+nd+2);
    int fff = 0;
    if(nd<2)fff=0;
    if(!fff){
        tb[last]--;dd[++nd]=last;
        dfs(dep+1,last,deal+1);
        tb[last]++;nd--;
    }
    }
    if(tb[last]==2&&(!flag)){
    tb[last]-=2;flag = 1;what = last;
    dfs(dep+1,last+1,deal+2);
    flag = 0;tb[last]+=2;
    }
    if(last%10==8||last%10==9||last/10==3)return;
    if((bool)tb[last]+(bool)tb[last+1]+(bool)tb[last+2]==2 && (!flag)){
    if(!tb[last+1]){
        tb[last]--;tb[last+2]--;flag = 1;what=last+1;
        dfs(dep+1,last,deal+2);
        tb[last]++;tb[last+2]++;flag = 0;
    }
    if(!tb[last+2]){
        tb[last]--;tb[last+1]--;flag = 1;what = last+2;
        dfs(dep+1,last,deal+2);
        tb[last]++;tb[last+1]++;flag = 0;
    }
    }
}

void work(){
    flag = 0;num=0;
    int fg = contry_no_double();
    if(fg)return;
    flag = 0;
    seven_double();
    dfs(1,1,0);
    int a1=0;
    for(int i=1;i<=37;i++){
    if(i%10==0)continue;
    if(ans[i]&&tb[i]==4)ans[i]=0;
    if(ans[i])a1++;
    }
    if(a1==0){puts("Nooten");}
    else{
    printf("%d ",a1);
    for(int i=1;i<=9;i++) if(ans[i])printf("%dm ",i);
    for(int i=1;i<=9;i++) if(ans[10+i])printf("%ds ",i);
    for(int i=1;i<=9;i++) if(ans[20+i])printf("%dp ",i);
    for(int i=1;i<=7;i++) if(ans[30+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 0;
}