FFT模版题。
观察题目，我们可以发现，只要把序列b倒过来，再联想一下乘法运算。。。
我们会发现，将序列a和序列b当作100进制数，做一次乘法，然后从低到高每一位便是答案了（乘完无需进位）
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cctype>
#include <complex>
#define rep(i, l, r) for(int i=l; i<=r; i++)
#define down(i, l, r) for(int i=l; i>=r; i--)
#define maxn 400009
#define cd complex <double>
#define PI acos(0.0)*2.0
#define ll long long
using namespace std;
inline int read()
{int x=0, f=1; char ch=getchar();while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}while (isdigit(ch)) x=x*10+ch-'0', ch=getchar();return x*f;
}
cd a[maxn], b[maxn], c[maxn], A[maxn];
int n, m, n1[maxn], n2[maxn], s[maxn];
int ans[maxn];void fft(cd *a, bool flag)
{rep(i, 0, n-1) s[i]=0;for(int i=1, j=n; i<n; i*=2, j/=2) rep(h, j/2, j-1) s[h]+=i;for(int i=1; i<n; i*=2) rep(j, 0, i-1) s[j+i]+=s[j];rep(i, 0, n-1) A[i]=a[s[i]];double pi=flag?PI:-PI;for(int step=1; step<n; step*=2){cd e=exp(cd(0, 2.0*pi/double(step*2))), w=cd(1, 0);for (int pos=0; pos<step; pos++, w*=e) for(int i=pos; i<n; i+=step*2){cd ret=A[i], rec=w*A[i+step];A[i]=ret+rec, A[i+step]=ret-rec;}}if (!flag) rep(i, 0, n-1) A[i]/=n;rep(i, 0, n-1) a[i]=A[i];
}int main()
{m=read(); rep(i, 1, m) n1[m-i]=read(), n2[i-1]=read();n=1; while (n<m*2) n*=2;rep(i, 0, n-1) a[i]=cd(n1[i], 0); fft(a, true);rep(i, 0, n-1) b[i]=cd(n2[i], 0); fft(b, true);rep(i, 0, n-1) c[i]=a[i]*b[i]; fft(c, false);rep(i, 0, m-1) ans[i]=c[i].real()+0.5;down(i, m-1, 0) printf("%d\n", ans[i]);
}
