ZOJ，zoj題目分類

ZOJ，zoj題目分類

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:

1. Insert. You can insert a non-zero digit or an asterisk anywhere. For example, if you insert a "1" at the beginning of "2*3*4", the string becomes "12*3*4".
2. Swap. You can swap any two characters in the string. For example, if you swap the last two characters of "12*3*4", the string becomes "12*34*".

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 *".

Input

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.

Output

For each test case, output the minimal number of operations to make the given string able to represent a valid RPN.

Sample Input

31*111*234***

Sample Output

102

題意：給你插入和任意交換字串兩個字母的兩個操作，求最少的步數使得字串變成尾碼運算式，

思路：發現顯然我們只有缺少數位時候才用到插入，不然就是交換了，把*都交換到後面，而一個*都需要兩個數字。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 1010;char str[maxn];int ind[maxn];int main() {int t;scanf("%d", &t);while (t--) {scanf("%s", str);int op = 0, num = 0;for (int i = 0; str[i]; i++) {if (str[i] == '*')op++;else num++;}if (num == strlen(str)) {printf("0\n");continue;}int ans = 0;if (op >= num) {ans += op - num + 1;num = op - num + 1;op = 0;}else {op = 0;num = 0;}int cnt = 0;for (int i = strlen(str)-1; i >= 0; i--)if (str[i] != '*')ind[cnt++] = i;cnt = 0;for (int i = 0; str[i]; i++) {if (str[i] != '*')num++;else op++;if (op >= num) {swap(str[i], str[ind[cnt++]]);ans++;op--;num++;}}if (str[strlen(str)-1] != '*')ans++;printf("%d\n", ans);}return 0;}

ZOJ 1049 助

你的程式有一個問題：
在定義前是不能有其它語句的，所以這一點你就錯了。
修改下如下：
#include<stdio.h>
#include<math.h>
int main()
{
int nCases,i;
int n;
double x,y,pi=3.14159 ;//定義語句必須放在最前

scanf("%d",&nCases);
for (i=1;i<=nCases;i++)
{

scanf("%lf %lf",&x,&y); //雙精確度的輸入應為%lf
n=ceil(pi*(x*x+y*y)/100);
printf(" Property %d: This property will begin eroding in year %d.\n",i,n);
}
printf(" END OF OUTPUT. \n");
return 0;
}
 
ZOJ 1049 助

你的程式有一個問題：
在定義前是不能有其它語句的，所以這一點你就錯了。
修改下如下：
#include<stdio.h>
#include<math.h>
int main()
{
int nCases,i;
int n;
double x,y,pi=3.14159 ;//定義語句必須放在最前

scanf("%d",&nCases);
for (i=1;i<=nCases;i++)
{

scanf("%lf %lf",&x,&y); //雙精確度的輸入應為%lf
n=ceil(pi*(x*x+y*y)/100);
printf(" Property %d: This property will begin eroding in year %d.\n",i,n);
}
printf(" END OF OUTPUT. \n");
return 0;
}