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
",i);return 0;}
		sum+=now/100;now%=100;
	}
	printf("%d
",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
",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-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