题解：需要将所有的前n次变换存储到对应的数组里，在第i-j次变换就直接数组相减即可。

//  #pragma GCC optimize(2, 3, "Ofast", "inline")
#include <bits/stdc++.h>
// #define endl '\n'
using namespace std;
double k[100010];
double theta[100010];
void work()
{int n,m;scanf("%d%d",&n,&m);memset(k,1,sizeof(k));k[0]=1;theta[0]=0;int type;double oper;for(int i=1;i<=n;i++){scanf("%d,%lf",&type,&oper);if(type==1){k[i]=k[i-1]*oper;theta[i]=theta[i-1];}else{k[i]=k[i-1];theta[i]=theta[i-1]+oper;}}int i,j;double x,y;double deltak;double deltatheta;for(int q=0;q<m;q++){scanf("%d %d %lf %lf",&i,&j,&x,&y);deltak=k[j]/k[i-1];deltatheta=theta[j]-theta[i-1];double x1=x*deltak;double y1=y*deltak;x=x1*cos(deltatheta)-y1*sin(deltatheta);y=x1*sin(deltatheta)+y1*cos(deltatheta);// cout<<x<<" "<<y<<endl;printf("%.3f %.3f\n",x,y);}
}
int main()
{// ios::sync_with_stdio(false);// cin.tie(0);// cout.tie(0);work();return 0;
}