牛客算法周周练11水题ABIN Liusy的博客-

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第一次练习赛AK。

A 切题之路

纯模拟，没什么好说的。

#include <bits/stdc++.h> #pragma GCC optimize(2) #pragma GCC optimize(3) typedef unsigned long long ll; const ll maxn=1e6; using namespace std; namespace IO{ char ibuf[1<<21],*ip=ibuf,*ip_=ibuf; char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21); inline char gc(){ if(ip!=ip_)return *ip++;         ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin); return ip==ip_?EOF:*ip++; } inline void pc(char c){ if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf; *op++=c; } inline ll read(){ register ll x=0,ch=gc(),w=1; for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1; for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48; return w*x; } template<class I> inline void write(I x){ if(x<0)pc('-'),x=-x; if(x>9)write(x/10);pc(x%10+'0'); } class flusher_{ public: ~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);} }IO_flusher; } using namespace IO; ll n,t,a,b,p[maxn],q[maxn],sum1,sum2,ans1,ans2; int main() {     n=read();t=read();a=read();b=read();sum1=sum2=t; for(int i=1;i<=n;i++) p[i]=read();//时间 for(int i=1;i<=n;i++) q[i]=read();//难度 for(int i=1;i<=n;i++) { if(q[i]<a&&sum1>=p[i])sum1-=p[i],ans1++; if(q[i]<b&&sum2>=p[i])sum2-=p[i],ans2++; else if(q[i]>=b&&sum2>=2*p[i])sum2-=2*p[i],ans2++; } write(ans1);pc(32);write(ans2); return 0; } 

C 红球进黑洞

感觉大佬们好像是被题面吓到了。。其实就是单纯的模拟。
差点超时，不是最佳解。

#include <bits/stdc++.h> #pragma GCC optimize(2) #pragma GCC optimize(3) typedef unsigned long long ll; const ll maxn=1e6; using namespace std; namespace IO{ char ibuf[1<<21],*ip=ibuf,*ip_=ibuf; char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21); inline char gc(){ if(ip!=ip_)return *ip++;         ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin); return ip==ip_?EOF:*ip++; } inline void pc(char c){ if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf; *op++=c; } inline ll read(){ register ll x=0,ch=gc(),w=1; for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1; for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48; return w*x; } template<class I> inline void write(I x){ if(x<0)pc('-'),x=-x; if(x>9)write(x/10);pc(x%10+'0'); } class flusher_{ public: ~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);} }IO_flusher; } using namespace IO; ll n,t,a[maxn],op1,l,r,k,sum; int main() {     n=read();t=read(); for(ll i=1;i<=n;i++) a[i]=read(); while(t--){         op1=read();sum=0; if(op1==1){             l=read();r=read(); for(ll i=l;i<=r;i++) sum+=a[i]; write(sum);pc('n'); } else{             l=read();r=read();k=read(); for(ll i=l;i<=r;i++) a[i]^=k; } } } 

D 游戏

正解是dfs+贪心。下面是菜鸡的偷鸡做法。
cin cout不能与IO的同用。

#include <bits/stdc++.h> #pragma GCC optimize(2) #pragma GCC optimize(3) typedef unsigned long long ll; const ll maxn=1e3+10; using namespace std; namespace IO{ char ibuf[1<<21],*ip=ibuf,*ip_=ibuf; char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21); inline char gc(){ if(ip!=ip_)return *ip++;         ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin); return ip==ip_?EOF:*ip++; } inline void pc(char c){ if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf; *op++=c; } inline ll read(){ register ll x=0,ch=gc(),w=1; for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1; for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48; return w*x; } template<class I> inline void write(I x){ if(x<0)pc('-'),x=-x; if(x>9)write(x/10);pc(x%10+'0'); } class flusher_{ public: ~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);} }IO_flusher; } //using namespace IO; char a[maxn][maxn]; int t,n,m; int main(){ scanf("%d",&t); while(t--){ scanf("%d%d",&n,&m); //n=read();m=read(); for(int i=1;i<=n;i++){             cin>>a[i]+1; } if(a[1][1]=='R') puts("dreagonm"); else if(a[1][1]=='G') puts("fengxunling"); else puts("BLUESKY007"); } } 

E 积木大赛

经典差分题，没啥好说的。。

#include <bits/stdc++.h> #pragma GCC optimize(2) #pragma GCC optimize(3) typedef long long ll; const ll maxn=1e6; using namespace std; namespace IO{ char ibuf[1<<21],*ip=ibuf,*ip_=ibuf; char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21); inline char gc(){ if(ip!=ip_)return *ip++;         ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin); return ip==ip_?EOF:*ip++; } inline void pc(char c){ if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf; *op++=c; } inline ll read(){ register ll x=0,ch=gc(),w=1; for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1; for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48; return w*x; } template<class I> inline void write(I x){ if(x<0)pc('-'),x=-x; if(x>9)write(x/10);pc(x%10+'0'); } class flusher_{ public: ~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);} }IO_flusher; } using namespace IO; ll n,a[maxn],b[maxn],ans1,ans2; int main() {     n=read(); for(int i=1;i<=n;i++){         a[i]=read();         b[i]=a[i]-a[i-1]; if(b[i]>0) ans1+=b[i]; else ans2-=b[i]; } write(max(ans1,ans2)); pc('n'); } 

完结。