还是DP题写起来容易的说。。。
直接看zky的一图流好了（话说zky用的是三分？。。。给跪）
1 /************************************************************** 2 Problem: 3437 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:1616 ms 7 Memory:32056 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 12 #define L q[l] 13 using namespace std; 14 typedef long long ll; 15 const int N = 1000005; 16 17 int n, l, r; 18 ll f[N], g[N], q[N]; 19 ll sum1[N], sum2; 20 21 inline int read() { 22 int x = 0; 23 char ch = getchar(); 24 while (ch < '0' || '9' < ch) 25 ch = getchar(); 26 while ('0' <= ch && ch <= '9') { 27 x = x * 10 + ch - '0'; 28 ch = getchar(); 29 } 30 return x; 31 } 32 33 inline ll calc1(int x, int y) { 34 return (ll) g[x] - g[y]; 35 } 36 37 inline ll calc2(int x, int y) { 38 return (ll) sum1[x] - sum1[y]; 39 } 40 41 inline bool pop_head(int x, int i) { 42 return calc1(q[x + 1], q[x]) < i * calc2(q[x + 1], q[x]); 43 } 44 45 inline bool pop_tail(int x, int i) { 46 return calc1(i, q[x]) * calc2(q[x], q[x - 1]) < calc1(q[x], q[x - 1]) * calc2(i, q[x]); 47 } 48 49 int main() { 50 int i, b; 51 n = read(); 52 for (i = 1; i <= n; ++i) 53 f[i] = read(); 54 for (l = r = 0, i = 1; i <= n; ++i) { 55 b = read(); 56 sum1[i] = sum1[i - 1] + b; 57 sum2 += (ll) i * b; 58 59 while (l < r && pop_head(l, i)) ++l; 60 f[i] += g[L] + i * (sum1[i] - sum1[L]) - sum2; 61 g[i] = f[i] + sum2; 62 while (l < r && pop_tail(r, i)) --r; 63 q[++r] = i; 64 } 65 printf("%lld ", f[n]); 66 return 0; 67 }