\(D1T1\) 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
