T1:
Code:
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define MN 100005 5 using namespace std; 6 inline int in(){ 7 int x=0;bool f=0; char c; 8 for (;(c=getchar())<'0'||c>'9';f=c=='-'); 9 for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0'); 10 return f?-x:x; 11 } 12 struct edge{ 13 int to,next; 14 }e[MN]; 15 int head[MN],edi[MN],t[MN],x[MN],cnt=0,n,tail=0,cur; 16 char op[MN]; 17 inline void ins(int x,int y){ 18 e[++cnt].to=y;e[cnt].next=head[x];head[x]=cnt; 19 } 20 void dfs(int p){ 21 if (op[p]=='T') t[++tail]=x[p]; 22 if (op[p]=='Q') x[p]=t[x[p]]; 23 for (int i=head[p];i;i=e[i].next) dfs(e[i].to); 24 if (op[p]=='T') --tail; 25 } 26 int main() 27 { 28 n=in();cur=0;for (int i=1;i<=n;++i){ 29 do op[i]=getchar();while (op[i]!='T'&&op[i]!='U'&&op[i]!='Q'); 30 if (op[i]=='T') getchar(),x[i]=getchar();else x[i]=in(); 31 if (op[i]=='U') ins(edi[cur-x[i]],i);else ins(i-1,i); 32 if (op[i]=='T'||op[i]=='U') edi[++cur]=i; 33 }dfs(0); 34 for (int i=1;i<=n;++i) if (op[i]=='Q') printf("%c ",(char)x[i]);return 0; 35 }
T2:动态规划。考虑将1-n从小到大插入数列。
设f[i][j]表示数列中已插入1-i,恰有j个"<"的方案数，则f[i][j]=f[i-1][j-1]*(i-j)+f[i-1][j]*(j+1).
Code:
1 #include<iostream> 2 #include<cmath> 3 #include<cstdio> 4 #include<cstdlib> 5 #include<cstring> 6 #include<algorithm> 7 #define MOD 2012 8 using namespace std; 9 inline int in(){ 10 int x=0;bool f=0; char c; 11 for (;(c=getchar())<'0'||c>'9';f=c=='-'); 12 for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0'); 13 return f?-x:x; 14 }int f[1003][1003],n,k; 15 int main() 16 { 17 n=in();k=in(); 18 for (int i=1;i<=n;++i) f[i][0]=f[i][i-1]=1; 19 for (int i=2;i<=n;++i) 20 for (int j=1;j<=min(k,i-1);++j) 21 f[i][j]=(f[i-1][j]*(j+1)%MOD+f[i-1][j-1]*(i-j)%MOD)%MOD; 22 printf("%d",f[n][k]);return 0; 23 }
T3:
考虑从后往前处理，可以发现是否开采或维修只对在其后星球的开采收入产生影响，即其后星球的最优方案与是否在该星球开采或修复无关。所以可以贪心处理本题。
Code:
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define MN 100005 5 using namespace std; 6 inline int in(){ 7 int x=0;bool f=0; char c; 8 for (;(c=getchar())<'0'||c>'9';f=c=='-'); 9 for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0'); 10 return f?-x:x; 11 } 12 int n,w,x[MN]; 13 double k,c,ans; 14 bool tp[MN]; 15 int main() 16 { 17 n=in();k=1-(in()/100.0);c=1+(in()/100.0);w=in(); 18 for (int i=1;i<=n;++i) tp[i]=in()-1,x[i]=in();ans=0.0; 19 for (int i=n;i;--i){ 20 if (!tp[i]) ans=max(ans,ans*k+x[i]); 21 else ans=max(ans,ans*c-x[i]); 22 }printf("%.2lf",ans*w); 23 return 0; 24 }