先把代码读懂，可以理解成为c是b的高维前缀和 a[i] 是第 i 维的长度 显然长度大于1的维数不会超过log2(m) 暴力还原回b就行了
#include <bits/stdc++.h> #define INF 2147483647 #define N 1000050 using namespace std; typedef long long LL; LL a[10050],c[N],sum[N]; int n,m; inline LL rd() { LL x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9') {if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();} return x*f; } int main() { n = rd(); printf("%d\n",n); for (int i=0;i<n;i++) { LL v = rd(); printf("%lld%c",v,i==n-1?'\n':' '); v==1 ? n--,i-- : a[i] = v; } m = rd(), a[n++] = INF; for (int i=0;i<m;i++) c[i] = rd(); sum[0] = 1; for (int i=1;i<n;i++) { sum[i] = 1LL * a[i-1] * sum[i-1]; if (sum[i] > m) { n = i; break; } } for (int j=0;j<n;j++) { int x = m % sum[j], y = m / sum[j] % a[j]; for (int i=m-1;i>=0;i--) { if (x) --x; else { x = sum[j] - 1; if (y) --y; else y = a[j] - 1; } if (y) c[i] -= ( c[i-sum[j]] ); } } printf("%d\n",m); for (int i=0;i<m;i++) printf("%lld%c",c[i],i==m-1?'\n':' '); return 0; } 