P2129 L国的战斗续之多路出击

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=500005,INF=2000000000;
 4 int X[maxn],Y[maxn],a[maxn],b[maxn];
 5 char cmd[maxn];
 6 int n,M,p=0,q=0,curx=1,cury=1;
 7 char c;
 8 int main() {
 9     cin>>n>>M;
10     for(int i=1; i<=n; i++) cin>>X[i]>>Y[i];
11     for(int i=0; i<M; i++) {
12         c=getchar();
13         while(c!=‘x‘&&c!=‘m‘&&c!=‘y‘) c=getchar();
14         cmd[i]=c;
15         if(c==‘m‘) cin>>a[i]>>b[i];
16     }
17     for(int i=M-1; i>=0; i--) {
18         if(cmd[i]==‘x‘) p=-p,curx=-curx;
19         else if(cmd[i]==‘y‘) q=-q,cury=-cury;
20         else p+=a[i],q+=b[i];
21     }
22     for(int i=1; i<=n; i++) printf("%d %d\n",X[i]*curx+p,Y[i]*cury+q);
23     return 0;
24 }