题意：略
一开始xjb贪心了好长时间...
然后发现可以从后往前枚举最晚时间，\(O(1)\)得到最小代价
确定最晚时间后就可以知道哪些可以用A啦！
一定要考虑这种变化变成不变的思想！
#include#include #include #include using namespace std;typedef long long ll;const int N = 1e5+5;inline int read() { char c=getchar(); int x=0,f=1; while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();} while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();} return x*f;}ll a, b, c, ans, now, big_cnt, big_sum, les_cnt, les_sum, s[N];int n, m, r[N], w[N], p, cnt_b, t[N];void solve() { ll cnt = 0; for(int i=1; i<=p; i++) cnt += t[i-1], s[i] = s[i-1] + cnt; ans = c >= 1e12 ? c : s[p] * c; for(int i=1; i<=p; i++) les_cnt += w[i-1], les_sum += les_cnt; while(p >= 2) { p--; // p+1 --> p big_cnt += w[p+1]; big_sum += big_cnt; les_sum -= les_cnt; les_cnt -= w[p]; if(c >= 1e12 && s[p]) continue; now = s[p] * c; //printf("hi %d %lld %lld %lld %lld %lld\n", p, s[p], big_cnt, big_sum, les_cnt, les_sum); if(b <= a) now += b * big_sum; else { ll k = big_sum - les_sum; if(k > 0) now += k * b + les_sum * a; else now += big_sum * a; } ans = min(ans, now); } printf("%lld", ans);}int main() { freopen("in", "r", stdin); scanf("%lld %lld %lld", &a, &b, &c); n=read(); m=read(); for(int i=1; i<=n; i++) t[read()]++; for(int i=1; i<=m; i++) r[i] = read(), w[r[i]]++, p = max(p, r[i]); solve();}