Do you know reverse Polish notation (RPN)? It is a known notation in the area of mathematics and computer science. It is also known as postfix notation since every operator in an expression follows all of its operands. Bob is a student in Marjar University. He is learning RPN recent days.
To clarify the syntax of RPN for those who haven't learnt it before, we will offer some examples here. For instance, to add 3 and 4, one would write "3 4 +" rather than "3 + 4". If there are multiple operations, the operator is given immediately after its second operand. The arithmetic expression written "3 - 4 + 5" in conventional notation would be written "3 4 - 5 +" in RPN: 4 is first subtracted from 3, and then 5 added to it. Another infix expression "5 + ((1 + 2) × 4) - 3" can be written down like this in RPN: "5 1 2 + 4 × + 3 -". An advantage of RPN is that it obviates the need for parentheses that are required by infix.
In this problem, we will use the asterisk "*" as the only operator and digits from "1" to "9" (without "0") as components of operands.
You are given an expression in reverse Polish notation. Unfortunately, all space characters are missing. That means the expression are concatenated into several long numeric sequence which are separated by asterisks. So you cannot distinguish the numbers from the given string.
You task is to check whether the given string can represent a valid RPN expression. If the given string cannot represent any valid RPN, please find out the minimal number of operations to make it valid. There are two types of operation to adjust the given string:
The strings "2*3*4" and "12*3*4" cannot represent any valid RPN, but the string "12*34*" can represent a valid RPN which is "1 2 * 34 *".
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
There is a non-empty string consists of asterisks and non-zero digits. The length of the string will not exceed 1000.
For each test case, output the minimal number of operations to make the given string able to represent a valid RPN.
3 1*1 11*234** *
1 0 2
題解及代碼:
比賽的時候坑了,思路有點漏洞。這道題目的意思就是給出一個後綴表達式,看它是否是一個合理的表達式,如果不合理,我們有兩種操作:1,任意交換兩個字符 2.添加一個數字或者*。其實是一道貪心的題目,我們會發現,如果有n個*的話,那麼表達式裡面至少要有n+1個數字,否則,我們就要加上一些數字使其滿足n+1,其次貪心的策略就是把加上的數字放在表達式的最前面,然後當判斷到某個*時,前面數字的數量少於2,那麼我們就從這個*後面找一個數字與其換位置,直到所有的*被算完。
#include#include #include using namespace std; char s[1010]; int main() { int cas; scanf("%d",&cas); while(cas--) { scanf("%s",s); int ans=0,len=strlen(s),p=-1,nstar=0,nnum=0; for(int i=0;i =nnum) { ans=(nstar-nnum+1); num=ans; } if(s[len-1]!='*') { ans++; swap(s[p],s[len-1]); } for(int i=0;i =2) num--; else { ans++; while(k>i) { if(s[k]!='*') { swap(s[i],s[k]); num++; break; } k--; } } } else num++; printf("%d\n",ans); } return 0; }