Jan.18th
注：黑粗体为超链接
大特判
简单DFS
两个指针记录一下Alice和Bob当前吃到哪里就行 Alice先吃
一开始看上去很不可做……然后发现数据范围非常非常小
那就DFS嘛 :-)
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#include<cstdio>
#include<cstring> #include<algorithm> #include<cctype> #include<ctime> #include<cstdlib> #include<string> #include<queue> #include<cmath> #define Rep(x,a,b) for (int x=a;x<=b;x++) #define Per(x,a,b) for (int x=a;x>=b;x--) #define ll long long using namespace std; inline int IN(){ int x=0,ch=getchar(),f=1; while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar(); if (ch=='-'){f=-1;ch=getchar();} while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} return x*f; } int a,b,n,hp[11],axl[11],cnt[11],Ans=1<<30; void Hit(int pos){ cnt[pos]++; hp[pos]-=a; hp[pos-1]-=b; hp[pos+1]-=b; } void DontHit(int pos){ cnt[pos]--; hp[pos]+=a; hp[pos-1]+=b; hp[pos+1]+=b; } bool AllKill(){ Rep(i,2,n) if (hp[i]>=0) return 0; return 1; } void DFS(int cs,int np){ if (AllKill()){ if (cs<Ans){ Ans=cs; memcpy(axl,cnt,sizeof cnt); } return; } Rep(i,np,n-1) if (hp[i]>=0||hp[i-1]>=0||hp[i+1]>=0){ Hit(i); DFS(cs+1,i); DontHit(i); } } int main(){ n=IN(),a=IN(),b=IN(); Rep(i,1,n) hp[i]=IN(); int nc=0; while (hp[1]>=0) nc++,Hit(2); while (hp[n]>=0) nc++,Hit(n-1); DFS(nc,2); printf("%d\n",Ans); Rep(i,2,n-1) while (axl[i]) axl[i]--,printf("%d ",i); } |
用两个单调队列维护就行
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#include<cstdio>
#include<cstring> #include<algorithm> #include<cctype> #include<ctime> #include<cstdlib> #include<string> #include<queue> #include<cmath> #define Rep(x,a,b) for (int x=a;x<=b;x++) #define Per(x,a,b) for (int x=a;x>=b;x--) #define ll long long using namespace std; inline int IN(){ int x=0,ch=getchar(),f=1; while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar(); if (ch=='-'){f=-1;ch=getchar();} while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} return x*f; } int n,k,big[100005],sml[100005],a[100005]; int bt,st,bh,sh,Ans[100005][2]; void Push(int val){ while (bt>=bh&&val>big[bt]) bt--;big[++bt]=val; while (st>=sh&&val<sml[st]) st--;sml[++st]=val; } void Pop(int val){ if (big[bh]==val) bh++; if (sml[sh]==val) sh++; } int main(){ n=IN(),k=IN(); Rep(i,1,n) a[i]=IN(); int l=1,Max=1,num=1; sh=bh=1; Push(a[1]); Ans[1][0]=Ans[1][1]=1; Rep(i,2,n){ Push(a[i]); while (big[bh]-sml[sh]>k) Pop(a[l]),l++; if (i-l+1>Max){ Max=i-l+1; num=1; Ans[1][0]=l,Ans[1][1]=i; } else if (i-l+1==Max){ num++; Ans[num][0]=l; Ans[num][1]=i; } } printf("%d %d\n",Max,num); Rep(i,1,num) printf("%d %d\n",Ans[i][0],Ans[i][1]); } |
暴力枚举每行每列涂不涂就行
标算是很复杂的东西……但这些我们不管
我们模拟退火就行 模拟退火是神器
晚上做了点528给初三爷们的作业……我去试了一下难度1
我这么弱 所以毫无疑问地爆炸了
我用了记忆化搜索
先存一下每个数字出现的次数
fi表示从1~i能获取的最大价值
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#include<cstdio>
#include<cstring> #include<algorithm> #include<cctype> #include<ctime> #include<cstdlib> #include<string> #include<queue> #include<cmath> #define Rep(x,a,b) for (int x=a;x<=b;x++) #define Per(x,a,b) for (int x=a;x>=b;x--) #define ll long long using namespace std; inline int IN(){ int x=0,ch=getchar(),f=1; while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar(); if (ch=='-'){f=-1;ch=getchar();} while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} return x*f; } ll f[100005]; int n,cnt[100005],Max; ll DFS(int val){ if (val<=0) return 0; if (~f[val]) return f[val]; f[val]=0; if (1ll*cnt[val]*val+DFS(val-2)>f[val]) f[val]=1ll*cnt[val]*val+DFS(val-2); if (DFS(val-1)>f[val]) f[val]=DFS(val-1); return f[val]; } int main(){ n=IN(); Rep(i,1,n){ int x=IN(); cnt[x]++; if (x>Max) Max=x; } memset(f,-1,sizeof f); printf("%I64d\n",DFS(Max)); } |
存一下每个数字在每个数列中出现的位置
然后枚举一下i和j，其中i是当前第一行要处理的位置,j是之前任意的位置
判断一下每个数列中aj是否都出现在ai前面就行
C++ Code
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#include<cstdio>
#include<cstring> #include<algorithm> #include<cctype> #include<ctime> #include<cstdlib> #include<string> #include<queue> #include<cmath> #define Rep(x,a,b) for (int x=a;x<=b;x++) #define Per(x,a,b) for (int x=a;x>=b;x--) #define ll long long using namespace std; inline int IN(){ int x=0,ch=getchar(),f=1; while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar(); if (ch=='-'){f=-1;ch=getchar();} while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} return x*f; } int n,k,a[5][1000],pos[5][1001],f[1000]; bool Judge(int form,int latt){ Rep(i,1,k-1) if (pos[i][form]>pos[i][latt]) return 0; return 1; } int main(){ n=IN(),k=IN(); Rep(i,0,k-1) Rep(j,0,n-1) a[i][j]=IN(),pos[i][a[i][j]]=j; Rep(i,0,n-1){ f[i]=1; Rep(j,0,i-1) if (f[j]+1>f[i]) if (Judge(a[0][j],a[0][i])) f[i]=f[j]+1; } int Ans=0; Rep(i,0,n-1) if (f[i]>Ans) Ans=f[i]; printf("%d\n",Ans); } |
记录一下每个位置最长连续上升序列的左端点和右端点
然后枚举断点就行……这种题我WA了9发 凸(艹皿艹 )
Jan 27, 2024 05:32:55 PM
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