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search.xml
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search.xml
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<?xml version="1.0" encoding="utf-8"?>
<search>
<entry>
<title><![CDATA[UEFI和传统引导windows的一点区别]]></title>
<url>%2F2019%2F11%2F09%2FUEFI%E5%92%8C%E4%BC%A0%E7%BB%9F%E5%BC%95%E5%AF%BCwindows%E7%9A%84%E4%B8%80%E7%82%B9%E5%8C%BA%E5%88%AB%2F</url>
<content type="text"><![CDATA[winload.exe 和winload.efi 在Windows传统引导中 启动文件项目的值是\Windows\system32\winload.exe 在Windows UEFI 引导中 启动文件项目的值是\Windows\system32\winload.efi 这个问题是我今天在使用VHDX来启动系统的时候遇到的，如果是exe的话就会卡住无法进入系统]]></content>
<tags>
<tag>Windows</tag>
</tags>
</entry>
<entry>
<title><![CDATA[Kruskal最小生成树模板]]></title>
<url>%2F2019%2F11%2F08%2FKruskal%E6%9C%80%E5%B0%8F%E7%94%9F%E6%88%90%E6%A0%91%E6%A8%A1%E6%9D%BF%2F</url>
<content type="text"><![CDATA[Kruskal最小生成树模板 12345678910111213141516171819202122232425262728293031323334#include<bits/stdc++.h>using namespace std;int n,m,f[5001],ans,t;struct s{ int u,v,w;}e[200001];bool cmp(s a, s b){ return a.w<b.w;}int find(int find_x){ return (f[find_x]==find_x)?(find_x):(f[find_x]=find(f[find_x]));}int main(){ cin>>n>>m; for(int i=0;i<=n;i++) f[i]=i; for(int i=0;i<m;i++) cin>>e[i].u>>e[i].v>>e[i].w; sort(e,e+m,cmp); for(int i=0;i<m;i++){ int a=find(e[i].u); int b=find(e[i].v); if(f[a]!=f[b]){ f[a]=b; ans+=e[i].w; t++; } } if(t==n-1) cout<<ans; else cout<<"orz"; return 0;}]]></content>
<tags>
<tag>luogu</tag>
<tag>最小生成树</tag>
<tag>kruskal</tag>
</tags>
</entry>
<entry>
<title><![CDATA[P1641 [SCOI2010]生成字符串]]></title>
<url>%2F2019%2F07%2F28%2FP1641-SCOI2010-%E7%94%9F%E6%88%90%E5%AD%97%E7%AC%A6%E4%B8%B2%2F</url>
<content type="text"><![CDATA[题意是n个1，m个0，问有多少种组合使在前k个数中1不少于0 1234567891011121314151617181920212223242526272829303132333435363738#include<bits/stdc++.h>using namespace std;typedef unsigned long long ull;typedef unsigned int uint;typedef long long ll;#define infi 0x3f3f3f3f#define fi first#define se second#define mp make_pairconst int N=2e6+1,mod=20100403;int n,m,fact[N],inv[N];//inv表示N!的逆元 ，fact表示N!modp int qpow(int a, int b) {//快速幂 int res=1; while(b){ if(b&1)res=1ll*res*a%mod; a=1ll*a*a%mod; b>>=1; } return res;}int C(int x, int y) { int z=1ll*fact[x]*inv[y]%mod; return 1ll*z*inv[x-y]%mod;}// ans= c(n)(n+m)-c(n-1)(n+m) zcyint main(){ int ans,n,m,t=1,e=0; ios::sync_with_stdio(0); cin>>n>>m; fact[0]=1; for(int i=1;i<=n+m;i++) fact[i]=(1ll*fact[i-1]*i)%mod; inv[n+m]=qpow(fact[n+m],mod-2); for (int i=n+m-1;i>=0;i--) inv[i]=1ll*inv[i+1]*(i+1)%mod; cout<<(C(n+m,m)-C(n+m,m-1)+mod)%mod; return 0;}]]></content>
<tags>
<tag>luogu</tag>
<tag>乘法逆元</tag>
</tags>
</entry>
<entry>
<title><![CDATA[P1156 垃圾陷阱]]></title>
<url>%2F2019%2F07%2F25%2FP1156%2F</url>
<content type="text"><![CDATA[题意：d深，10能量，一小时能量减一，n个物品有投入时间，高度，能量的属性，问何时可以达到d深或最长时间 12345678910111213141516171819202122232425262728293031#include<bits/stdc++.h>using namespace std;struct data{ int t,f,h;}ct[2001];bool cm(data a,data b){ return a.t<b.t;}int dp[200]={10};//到i高度还有多少时间 int d,g,t,dnow;int main(){ ios::sync_with_stdio(0); cin>>d>>g; for(int i=1;i<=g;i++) cin>>ct[i].t>>ct[i].f>>ct[i].h; sort(ct+1,ct+g+1,cm); for(int i=1;i<=g;i++){ for(int j=d;j>=0;j--){ if(dp[j]>=ct[i].t){ if(j+ct[i].h>=d){ cout<<ct[i].t; return 0; } dp[j+ct[i].h]=max(dp[j+ct[i].h],dp[j]); dp[j]+=ct[i].f; } } } cout<<dp[0]; return 0;}]]></content>
<tags>
<tag>luogu</tag>
<tag>dp</tag>
</tags>
</entry>
<entry>
<title><![CDATA[test]]></title>
<url>%2F2019%2F07%2F15%2Ftest%2F</url>
<content type="text"><![CDATA[Test is a good day]]></content>
</entry>
<entry>
<title><![CDATA[Hello World]]></title>
<url>%2F2019%2F07%2F15%2Fhello-world%2F</url>
<content type="text"><![CDATA[Welcome to Hexo! This is your very first post. Check documentation for more info. If you get any problems when using Hexo, you can find the answer in troubleshooting or you can ask me on GitHub. Quick StartCreate a new post1$ hexo new "My New Post" More info: Writing Run server1$ hexo server More info: Server Generate static files1$ hexo generate More info: Generating Deploy to remote sites1$ hexo deploy More info: Deployment]]></content>
</entry>
</search>