寒假玩了十天了已经…(好罪恶
今天开始刷题吧….
dp[i][j]代表从1~i插入j个乘号能达到的最大值
例如找到1~i时要插入j个括号
这时我们应该在1~i中间找到一个乘法的前驱1~k和1~k经过j-1次乘法的最大值
结果为1~k经过j-1次乘法后的最大值*k~i的数值
#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<math.h>#include<queue>#include<stack>#include<string>#include<vector>#include<map>#include<set>using namespace std;#define rfor(i,a,b) for(i=a;i<=b;++i)#define lfor(i,a,b) for(i=a;i>=b;--i)#define sfor(i,a,h) for(i=h[a];i!=-1;i=e[i].next)#define mem(a,b) memset(a,b,sizeof(a))#define mec(a,b) memcpy(a,b,sizeof(b))#define cheak(i) printf("%d ",i)#define min(a,b) (a>b?b:a)#define max(a,b) (a>b?a:b)#define inf 0x3f3f3f3f#define lowbit(x) (x&(-x))typedef long long LL;#define maxn 50#define maxm maxn*maxn#define lson(x) (splay[x].son[0])#define rson(x) (splay[x].son[1])LL dp[maxn][maxn],A[maxn][maxn];char str[maxn];int main(){ int i,j,k,T,m; scanf("%d",&T); while(T--) { scanf("%s",str+1); scanf("%d",&m); int len=strlen(str+1); rfor(i,1,len) { A[i][i]=str[i]-'0'; rfor(j,i+1,len) { A[i][j]=A[i][j-1]*10+str[j]-'0'; } } mem(dp,0); rfor(i,1,len) dp[i][1]=A[1][i]; rfor(i,2,m) { rfor(j,i,len) { rfor(k,1,j-1) { dp[j][i]=max(dp[j][i],dp[k][i-1]*A[k+1][j]); } } } /*rfor(i,1,m) { rfor(j,1,len) printf("%lld ",dp[j][i]);printf("n"); }*/ printf("%lldn",dp[len][m]); } return 0;}