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如何保證訊息佇列的可靠性傳輸?

板子

const int N=1e6;
struct FFT{
  const double pi=acos(-1);
  struct cp{
      double x,y;
      cp operator +(const cp &o)const{return (cp){x+o.x,y+o.y};}
      cp operator -(const cp &o)const{return (cp){x-o.x,y-o.y};}
      cp operator *(const cp &o)const{return (cp){x*o.x-y*o.y,y*o.x+x*o.y};}
      cp 
operator *(const double &o)const{return (cp){x*o,y*o};} cp operator !()const{return (cp){x,-y};} }a[N],b[N],c[N],w[N]; int l,r[N]; int n,m; void fft_init() { l=0; for (n=1;n<=m;n<<=1) l++; } void fft_clear() { rep(i,0,n) a[i].x=a[i].y=b[i].x=b[i].y=0; }
void fft(cp x[],int k,int v){ for(int i=0,j=0;i<k;i++) { if(i>j)std::swap(x[i],x[j]); for(int l=k>>1;(j^=l)<l;l>>=1); } w[0]=(cp){1,0}; for(int i=2;i<=k;i<<=1){ cp g=(cp){cos(2*pi/i),(v?-1:1)*sin(2*pi/i)};
for(int j=(i>>1);j>=0;j-=2)w[j]=w[j>>1]; for(int j=1;j<i>>1;j+=2)w[j]=w[j-1]*g; for(int j=0;j<k;j+=i){ cp *a=x+j,*b=a+(i>>1); for(int l=0;l<i>>1;l++){ cp o=b[l]*w[l]; b[l]=a[l]-o; a[l]=a[l]+o; } } } if(v)for(int i=0;i<k;i++)x[i]=(cp){x[i].x/k,x[i].y/k}; } IL void get_cj(ll *A,ll *B,ll *C,int len) { rep(i,0,len) (i&1?a[i>>1].y:a[i>>1].x)=A[i]; rep(i,0,len) (i&1?b[i>>1].y:b[i>>1].x)=B[i]; m=len; fft_init(); fft(a,n,0); fft(b,n,0); for(int i=0;i<n;i++){ int j=(n-1)&(n-i); c[i]=(a[i]*b[i]*4-(a[i]-!a[j])*(b[i]-!b[j])*(((i&n>>1)?(cp){1,0}-w[i^n>>1]:w[i]+(cp){1,0})))*0.25; } fft(c,n,1); rep(i,0,len) if (i&1) C[i]=(ll)(c[i>>1].y+0.1); else C[i]=(ll)(c[i>>1].x+0.1); fft_clear(); } }F;