T1 ：最小值（min）题解 ——2019.10.15

思路：

对于 % 30 的数据，可以想到一个 Dp 方程：

其中dp[i]表示分割[1,i]的最大答案

代码：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 const int Nt = ; const long long inf = 0x3f3f3f3f3f3f3f3f;
 int ri() {
     char c = getchar(); int x = , f = ; for(;c < '' || c > ''; c = getchar()) if(c == '-') f = -;
     for(;c >= '' && c <= ''; c = getchar()) x = (x << ) + (x << ) - '' + c; return x * f;
 }
 int a[Nt], st[Nt], tp, n, A, B, C, D;
 long long T[Nt << ], f[Nt], mx[Nt];

 void Up(int i, long long x)
 {for(T[i += Nt] = x; i >>= ;) T[i] = std::max(T[i << ], T[i <<  | ]);}

 long long Cal(long long x) {return ((A * x + B) * x + C) * x + D;}

 int main() {
     freopen("min.in","r",stdin);
     freopen("min.out","w",stdout);
     n = ri(); A = ri(); B = ri(); C = ri(); D = ri();
     for(int i = ;i <= n; ++i) a[i] = ri();
     std::memset(T, -0x3f, sizeof(T));
     f[] = ; mx[] = ; st[tp = ] = a[]; Up(, Cal(a[]));
     for(int i = ;i <= n; ++i) {
         f[i] = T[]; long long x = f[i];
         for(;st[tp] > a[i + ] && tp;) x = std::max(x, mx[tp]), Up(tp--, -inf);
         st[++tp] = a[i + ]; mx[tp] = x; Up(tp, x + Cal(st[tp]));
     }
     printf("%lld\n", f[n]);
     return ;
 }