状态不好！！……
prob1&prob2，傻逼题，都是基础操作的练习，练手的，一个练循环，一个练\(sort\)和循环，垃圾，上代码：
\(T1\):
#include<iostream>
#include<cstdio>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;i++)
#define int long long
inline int read()
{
int x=0,f=1;char ch=getchar();
for(;!isalnum(ch);ch=getchar()) if(ch=='-') f=-1;
for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
return x*f;
}
int ans[2];
signed main()
{
int t=in;
while(t--)
{
int n=in;
ans[0]=ans[1]=0;
fur(i,1,n) ans[i%2]+=in;
fur(i,1,n) ans[(i%2)^1]+=in;
printf("%lld\n",ans[0]>ans[1]?ans[1]:ans[0]);
}
return 0;
}#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;i++)
#define jiba signed
const int xx=1e5+10;
inline int read()
{
int x=0,f=1;char ch=getchar();
for(;!isalnum(ch);ch=getchar()) if(ch=='-') f=-1;
for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
return x*f;
}
int a[xx];
jiba main()
{
int t=in;
while(t--)
{
int n=in;
fur(i,1,n) a[i]=in;
sort(a+1,a+n+1);
if(a[1]<a[2]-1)
{
printf("%d\n",a[1]);
continue;
}
if(a[n]>a[n-1]+1)
{
printf("%d\n",a[n]);
continue;
}
fur(i,1,n)
{
if(a[i]==a[i+1])
{
printf("%d\n",a[i]);
break;
}
}
}
return 0;
}prob5:Dish Owner
傻逼并查集，结果还调了老子半天，\(fuck\)，不说了，上代码：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;i++)
#define ll long long
inline int read()
{
int x=0,f=1;char ch=getchar();
for(;!isalnum(ch);ch=getchar()) if(ch=='-') f=-1;
for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
return x*f;
}
const int xx=1e4+10;
int fa[xx],s[xx];
inline int find(int i){return i==fa[i]?i:(fa[i]=find(fa[i]));}
int main()
{
int t=in;
while(t--)
{
int n=in;
fur(i,1,n) fa[i]=i,s[i]=in;
int m=in;
while(m--)
{
int op=in;
if(op==0)
{
int x=in,y=in;
int u=find(x),v=find(y);
if(u==v)
{
puts("Invalid query!");
continue;
}
if(s[u]>s[v]) fa[v]=u;
else if(s[u]<s[v]) fa[u]=v;
}
else
{
int x=in;
printf("%d\n",find(x));
}
}
}
return 0;
}prob3:Cooking Schedule
傻逼二分，二分最长长度，判断可否，特判一即可，注意当最长长度为\(k\)时在一段长度为\(len\)的\(0\)或\(1\)中的最优划分策略为修改\(len/(k+1)\)天:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;++i)
#define jiba signed
inline int read()
{
int x=0,f=1;char ch=getchar();
for(;!isalnum(ch);ch=getchar()) if(ch=='-') f=-1;
for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
return x*f;
}
const int xx=1e6+10;
int len[xx],all,need,ans1,ans2;
char op[xx];
inline bool check(int k)
{
int now=0;
fur(i,1,all)
{
if(len[i]>k)
{
int u=len[i];
now+=u/(k+1);
if(now>need) return false;
}
}
return true;
}
jiba main()
{
int t=in;
while(t--)
{
fur(i,1,all) len[i]=0;
int n=in,hd=2,tl=0;
all=0;need=in;
ans1=0;ans2=0;
scanf("%s",op);
len[++all]++;
op[0]-='0';
if(op[0]!=0) ans1++;
else ans2++;
fur(i,1,n-1)
{
op[i]=op[i]-'0';
if(op[i-1]!=op[i]) ++all;
if(op[i]!=i%2) ans1++;
else ans2++;
++len[all];
tl=max(tl,len[all]);
}
if(ans1<=need||ans2<=need)
{
puts("1");
continue;
}
int ans=tl;
while(hd<=tl)
{
int mid=(hd+tl)>>1;
if(check(mid)) tl=mid-1,ans=mid;
else hd=mid+1;
}
printf("%d\n",ans);
}
return 0;
}prob6:Triplets
30分部分分为暴力，在此不介绍。
正解数学推式子，将给定的式子拆开：设目前扫到的\(B\)序列的数为\(Y_p\)，将\(A\)与\(C\)序列\(sort\)后，小于等于\(Y_p\)的为\(X_{1-n}\)与\(Z_{1-m}\)，则我们要求的为\(\sum_{i=1}^{n}\sum_{j=1}^{m}(X_i+Y_p)* (Z_j+Y_p)\).考虑化简（因式分解之合并同类项），先提公因式\((X_i+Y_p)\)，则式子化为\(\sum_{i=1}^{n}((X_i+Y_p)* (\sum_{j=1}^{m}Z_j+m* Y_p))\)，同理，最后化简为\((\sum_{i=1}^{n}X_i+n* Y_p)* (\sum_{j=1}^{m}Z_j+m* Y_p)\).于是，确定方案为\(sort\)，再扫，逐个\(Y\)进行求解。至此，游戏结束：
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;++i)
#define int long long
#define jiba signed
inline int read()
{
int x=0,f=1;char ch=getchar();
for(;!isalnum(ch);ch=getchar()) if(ch=='-') f=-1;
for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
return x*f;
}
const int xx=1e5+10;
const int mod=1e9+7;
int a[xx],c[xx],b[xx];
jiba main()
{
int t=in+1;
while(--t)
{
int n=in,m=in,r=in;
fur(i,1,n) a[i]=in;sort(a+1,a+n+1);
fur(j,1,m) b[j]=in;sort(b+1,b+m+1);
fur(k,1,r) c[k]=in;sort(c+1,c+r+1);
int tmp1=0,tmp2=0,ans=0;
for(int i=1,j=1,k=1;j<=m;++j)
{
while(a[i]<=b[j]&&i<=n) tmp1=(tmp1+a[i])%mod,++i;
while(c[k]<=b[j]&&k<=r) tmp2=(tmp2+c[k])%mod,++k;
ans=(ans+(tmp1+(i-1)*b[j]%mod)%mod*(tmp2+(k-1)*b[j]%mod)%mod)%mod;
}
printf("%lld\n",ans);
}
return 0;
}prob4:CodeChef-SUBREM
树型\(DP\)，数组\(f[i]\)和\(w[i]\)分别记录\(i\)子树中的最大收益与点权和
转移方程：\(f[i]=max(\sum f[son[i][j]],-w[i]-x)\)
目标：\(w[1]+f[1]\)
两遍\(dfs\)，游戏结束：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;i++)
#define int long long
#define jiba signed
inline int read()
{
int x=0,f=1;char ch=getchar();
for(;!isalnum(ch);ch=getchar()) if(ch=='-') f=-1;
for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
return x*f;
}
const int xx=1e5+10;
int f[xx],w[xx],x,fa[xx];
bool vis[xx];
vector<int>e[xx];
inline void dfs1(int g)
{
fur(i,0,(int)e[g].size()-1)
{
if(vis[e[g][i]]) continue;
vis[e[g][i]]=true;
fa[e[g][i]]=g;
dfs1(e[g][i]);
w[g]+=w[e[g][i]];
}
}
inline void dfs2(int g)
{
f[g]=0;
fur(i,0,(int)e[g].size()-1)
{
if(e[g][i]==fa[g]) continue;
dfs2(e[g][i]);
f[g]+=f[e[g][i]];
}
f[g]=max(f[g],-w[g]-x);
}
inline void init(int n)
{
fur(i,1,n)
{
vis[i]=false;
e[i].clear();
fa[i]=0;
}
}
jiba main()
{
int t=in+1;
while(--t)
{
int n=in;x=in;
fur(i,1,n) w[i]=in;
fur(i,1,n-1)
{
int x=in,y=in;
e[x].push_back(y);
e[y].push_back(x);
}
vis[1]=true;
dfs1(1);
dfs2(1);
printf("%lld\n",f[1]+w[1]);
init(n);
}
return 0;
}\(ps\):以后考试要调整好心态，而且最好少\(dabai\)
原文地址：https://www.cnblogs.com/ALANALLEN21LOVE28/p/11313008.html
时间： 2024-07-31 18:47:51