ACM-ICPC 2018 焦作赛区网络预赛 B. Mathematical Curse

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#pragma comment(linker, "/STACK:102400000,102400000")
#include<unordered_map>
#include<unordered_set>
#include<algorithm>
#include<iostream>
#include<fstream>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iomanip>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cctype>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define pt(a) cout<<a<<endl
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
typedef pair<int,int> PII;
const ll mod = 1e9+7;
const int N = 1e3+10;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};ll t,n,m,k,num[N],dpmin[N][6],dpmax[N][6];
char s[10];int main() {ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);cin>>t;while(t--) {mst(dpmin,0),mst(dpmax,0);cin>>n>>m>>k;rep(i,1,n) cin>>num[i];cin>>s+1;rep(i,0,n) dpmin[i][0]=dpmax[i][0]=k;rep(i,1,n) {rep(j,1,m) {if(i<j) continue;///特判if(s[j]=='+') {dpmin[i][j]=min(dpmin[i-1][j-1]+num[i],dpmax[i-1][j-1]+num[i]);dpmax[i][j]=max(dpmin[i-1][j-1]+num[i],dpmax[i-1][j-1]+num[i]);}else if(s[j]=='-') {dpmin[i][j]=min(dpmin[i-1][j-1]-num[i],dpmax[i-1][j-1]-num[i]);dpmax[i][j]=max(dpmin[i-1][j-1]-num[i],dpmax[i-1][j-1]-num[i]);}else if(s[j]=='*') {dpmin[i][j]=min(dpmin[i-1][j-1]*num[i],dpmax[i-1][j-1]*num[i]);dpmax[i][j]=max(dpmin[i-1][j-1]*num[i],dpmax[i-1][j-1]*num[i]);}else if(s[j]=='/') {dpmin[i][j]=min(dpmin[i-1][j-1]/num[i],dpmax[i-1][j-1]/num[i]);dpmax[i][j]=max(dpmin[i-1][j-1]/num[i],dpmax[i-1][j-1]/num[i]);}if(i==j) {continue;}///if(i>j) {///dpmin[i][j]=min(dpmin[i-1][j],dpmin[i][j]);dpmax[i][j]=max(dpmax[i-1][j],dpmax[i][j]);///}}}cout<<dpmax[n][m]<<endl;}return 0;
}