很容易能得到状态转移方程 dp[ i ] = min( dp[ j ] + b[ j ] * a[ i ] )， 然后斜率优化一下。

一直以为炸精度了， 忽然发现手贱把while 写成了if 。。。。

#include
     
      #define LL long long#define fi first#define se second#define mk make_pair#define PLL pair
      
       #define PLI pair
       
        #define PII pair
        
         #define SZ(x) ((int)x.size())#define ull unsigned long longusing namespace std;const int N = 4e5 + 7;const int inf = 0x3f3f3f3f;const LL INF = 0x3f3f3f3f3f3f3f3f;const int mod = 1e9 + 7;const double eps = 1e-8;int n, que[N], head, rear;LL a[N], b[N], dp[N];double calc(int k, int j) {    return 1.0 * (dp[j] - dp[k]) * (b[k] - b[j]);}int main() {    scanf("%d", &n);    for(int i = 1; i <= n; i++) scanf("%lld", &a[i]);    for(int i = 1; i <= n; i++) scanf("%lld", &b[i]);    dp[1] = 0;    head = 1, rear = 0;    que[++rear] = 1;    for(int i = 2; i <= n; i++) {        while(rear - head + 1 >= 2 && calc(que[head], que[head + 1]) < a[i]) head++;        dp[i] = dp[que[head]] + b[que[head]] * a[i];        while(rear - head + 1 >= 2  && calc(que[rear-1], que[rear]) >= calc(que[rear], i)) rear--;        que[++rear] = i;    }    printf("%lld\n", dp[n]);    return 0;}/**/