NOIP 2004 题解

NOIP2004好水啊，我这种蒟蒻都能AK……；

T1：
https://www.luogu.org/problem/show?pid=1089

淼，最简单的模拟

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int q[15],ans,sum,n=12;void solve(){    for(int i=1;i<=n;i++)        scanf("%d",&q[i]);    for(int i=1;i<=n;i++)    {        int f=sum+300-q[i],c=f%100;        if(f>=100) ans+=f-c;        sum=c;        if(sum<0) cout<<-i<<endl,exit(0);    }    cout<<sum+ans*1.2;    return;}int main(){    solve();    return 0;}

T2：

https://www.luogu.org/problem/show?pid=1090

淼，堆，来个手打；

#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>using namespace std;const int MAXN=100010;int h[MAXN],n,ans,sum,v1,v2;void read(int &a){    int ans=0;    char c=getchar();    while(c<'0'||c>'9') c=getchar();    while(c>='0' && c<='9') ans*=10,ans+=c-'0',c=getchar();    a=ans;    return;} void swap(int a,int b){    int t;    t=h[a],h[a]=h[b],h[b]=t;    return;}void heap(int i){    int t,flag=1;    while(i*2<=n && flag)    {        if(h[i]>h[i*2] && i*2<=n) t=i*2;            else t=i;                if(h[t]>h[i*2+1] && i*2+1<=n) t=i*2+1;                    if(i!=t) swap(i,t),i=t;                        else flag=0;    }    return;}void creat(){    for(int i=n/2;i>=1;i--)        heap(i);        return;}void push(int s){    int i=n+1,flag=1;    h[i]=s,n++;    if(i==1) return;    while(i!=1 && flag)    {        if(h[i]<h[i/2]) swap(i,i/2);            else flag=0;                i/=2;    }    return;}void pop() {    swap(1,n),n--,heap(1);    return;}int top() {    return h[1];}void solve(){    cin>>n;    for(int i=1;i<=n;i++)    read(h[i]);    cin>>n;    creat();    while(n!=1)    {        v1=top(),pop();        v2=top(),pop();        sum=0,sum=v1+v2;        ans+=sum;        push(sum);    }    cout<<ans;    return;}int main(){    solve();    return 0;}

T3

https://www.luogu.org/problem/show?pid=1091

淼，正反两边最长上升子序列，最后加1（中间的人加了两遍）

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int n,dp1[20001],dp2[20001],num[20001],ans=-1;void solve(){    cin>>n;    for(int i=1;i<=n;i++) scanf("%d",&num[i]),dp1[i]=dp2[i]=1;    for(int i=1;i<=n;i++)        for(int j=1;j<i;j++)             if(num[i]>num[j])                 dp1[i]=max(dp1[j]+1,dp1[i]);    for(int i=n;i>=1;i--)        for(int j=n;j>i;j--)            if(num[i]>num[j])                dp2[i]=max(dp2[j]+1,dp2[i]);    for(int i=1;i<=n;i++) ans=max(ans,dp1[i]+dp2[i]);    cout<<n-ans+1;}int main(){    solve();    return 0;}

T4

https://www.luogu.org/problem/show?pid=1092

个人认为这是2004NOIP唯一一个值得做的题；

题解：
http://blog.csdn.net/qq_36312502/article/details/78072903

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=10001;int n,a[4][MAXN],ans[MAXN],a1,a2,a3;string s1,s2,s3;bool can[MAXN];bool pd(int y)//优化2{    for(int i=y+1;i<=n;i++)        if(ans[a[1][i]]!=-1 && ans[a[2][i]]!=-1 && ans[a[3][i]]!=-1)            if((ans[a[1][i]]+ans[a[2][i]])%n!=ans[a[3][i]] && (ans[a[1][i]]+ans[a[2][i]]+1)%n!=ans[a[3][i]])                return false;    return true;}void dfs(int x,int y,int ge){    if(y>n && ge^1)    {        for(int i=1;i<=n;i++) cout<<ans[i]<<" ";        exit(0);    }    if(!pd(y)) return;    a1=ans[a[1][y]],a2=ans[a[2][y]],a3=ans[a[3][y]];    if(ans[a[x][y]]==-1 && x<=3)    {        if(x==3)//优化2        {            int tot=a1+a2+ge;            if(tot<n && !can[tot])            {                can[tot]=1,ans[a[3][y]]=tot;                dfs(1,y+1,0);                can[tot]=0,ans[a[3][y]]=-1;            }            else if(tot>=n && !can[(tot+n)%n])             {                can[(tot+n)%n]=1,ans[a[3][y]]=(tot+n)%n;                dfs(1,y+1,1);                can[(tot+n)%n]=0,ans[a[3][y]]=-1;            }            else return;        }        else for(int i=n-1;i>=0;i--)//优化3；        {            if(!can[i])            {                can[i]=1,ans[a[x][y]]=i;                dfs(x+1,y,ge);                can[i]=0,ans[a[x][y]]=-1;            }        }        return;    }    else     {        if(x>=4)        {            if((ge && a1+a2+1==a3) || (!ge && a1+a2==a3)) dfs(1,y+1,0);            else if((ge && (a1+a2+1)%n==a3) || (!ge && (a1+a2)%n==a3)) dfs(1,y+1,1);            else return;        }        else dfs(x+1,y,ge);    }    return;}void solve(){    scanf("%d",&n);    cin>>s1>>s2>>s3;    for(int i=1;i<=n;i++)     a[1][i]=s1[n-i]-'A'+1,a[2][i]=s2[n-i]-'A'+1,a[3][i]=s3[n-i]-'A'+1;    memset(ans,-1,sizeof(ans));    dfs(1,1,0);    return;}int main(){    solve();    return 0;}