CF552E 字符串 表达式求值
http://codeforces.com/contest/552/problem/E
Vanya is doing his maths homework. He has an expression of form , where x1, x2, ..., xn are digits from 1 to 9, and sign represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
The first line contains expression s (1 ≤ |s| ≤ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs + and * .
The number of signs * doesn't exceed 15.
In the first line print the maximum possible value of an expression.
3+5*7+8*4
303
2+3*5
25
3*4*5
60
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
/** CF552E 字符串 表达式求值 题目大意:给定一个字符串仅仅是做1~9之间的加法和乘法,如今在表达式中加上一对括号。问怎样加才干使表达式的值最大 解题思路:左括号必须在一个*的后面,右括号必须在一个*的前面,假设不是这样一定不是最优。有了这个结论,分成三部分算一下就能够了 */ #include <string.h> #include <iostream> #include <algorithm> #include <stdio.h> using namespace std; typedef long long LL; char s[5005]; int n,a[105]; LL get(int l,int r) { LL ans=0,tmp=1; for(int i=l;i<=r;i++) { if(isdigit(s[i])) { tmp*=s[i]-'0'; } if(s[i]=='+') { ans+=tmp; tmp=1; } } ans+=tmp; tmp=1; LL sum=0; for(int i=0;i<n;i++) { if(i==l) { tmp*=ans; i=r; } else { if(isdigit(s[i])) { tmp*=s[i]-'0'; } if(s[i]=='+') { sum+=tmp; tmp=1; } } } sum+=tmp,tmp=1; // printf(">>>%d %d %d %d\n",l,r,ans,sum); return sum; } int main() { scanf("%s",s); n=strlen(s); int k=0; a[k++]=0; for(int i=0;i<n;i++) { if(s[i]=='*') { a[k++]=i-1; a[k++]=i+1; } } if(a[k-1]!=n-1) a[k++]=n-1; LL maxx=0; for(int i=0;i<k;i++) { for(int j=i;j<k;j++) { LL t=get(a[i],a[j]); maxx=max(maxx,t); } } printf("%lld\n",maxx); return 0; }