NOIP2004提高组题解

\(D1T1\) 津津的储蓄计划 \((OK)\)

\(D1T2\) 合并果子 \((OK)\)

\(D1T3\) 合唱队形 \((OK)\)

\(D1T4\) 虫食算 \((OK)\)

除了\(T4\)之外都挺水的.

\(T1\)额,纯简单模拟.

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define ll long long
using namespace std;
inline int read(){
    int x=0,o=1;char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    if(ch=='-')o=-1,ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int main(){
    int now=0,sum=0;
    for(int i=1;i<=12;++i){
        now+=300;int ys=read();now-=ys;
        if(now<0){printf("-%d\n",i);return 0;}
        sum+=now/100;now%=100;
    }
    printf("%d\n",120*sum+now);
    return 0;
}

\(T2\)堆的入门题吧.

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define ll long long
using namespace std;
inline int read(){
    int x=0,o=1;char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    if(ch=='-')o=-1,ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*o;
}
priority_queue<int,vector<int>,greater<int> >q;
int main(){
    int n=read(),ans=0;
    for(int i=1;i<=n;++i)q.push(read());
    while(q.size()>1){
        int x=q.top();q.pop();
        int y=q.top();q.pop();
        ans+=x+y;q.push(x+y);
    }
    printf("%d\n",ans);
    return 0;
}

\(T3\)把合唱队形拆成两个部分来求,则两个部分都是满足单调性的,因为数据范围很小,直接设\(f1[i]\)表示以\(a[i]\)结尾最长单调上升子序列的长度,\(f2[i]\)表示以\(a[i]\)开头的最长单调下降子序列的长度.

统计答案的时候在\(f1[i]+f2[i]-1\)中取\(max\)即可.

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define ll long long
using namespace std;
inline int read(){
    int x=0,o=1;char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    if(ch=='-')o=-1,ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*o;
}
const int N=105;
int ans,a[N],f1[N],f2[N];
int main(){
    int n=read();for(int i=1;i<=n;++i)a[i]=read();
    for(int i=1;i<=n;++i)f1[i]=f2[i]=1;
    for(int i=2;i<=n;++i)
        for(int j=1;j<i;++j)if(a[i]>a[j])f1[i]=max(f1[i],f1[j]+1);
    for(int i=n-1;i>=1;--i)
        for(int j=i+1;j<=n;++j)if(a[i]>a[j])f2[i]=max(f2[i],f2[j]+1);
    for(int i=1;i<=n;++i)ans=max(ans,f1[i]+f2[i]-1);
    printf("%d\n",n-ans);
    return 0;
}

\(T4\)直接从低位向高位开始爆搜填数,一边填数一边剪枝即可.博客

NOIP2004提高组题解

原文：https://www.cnblogs.com/PPXppx/p/11823058.html