额。。 不理解这题为什么会是数学题 0.0 可能是输入中有很长的字符串需要处理,然后获得字符串的位数是一个需要技巧的过程? 不过我直接STL过的 。。。真心感觉STL很方便 但是也不能依赖STL额。。顺便总结一下输出的时候如果没要求最后不输出回车 最后一定要输出回车 否则就会WAWAWA 我在这栽了三次 = =
以下是题和AC代码
Secret Research |
At a certain laboratory results of secret research are thoroughly encrypted. A result of a single experiment is stored as an information of its completion:
`positive result', `negative result', `experiment failed' or `experiment not completed'
The encrypted result constitutes a string of digits S, which may take one of the following forms:
positive result S = 1 or S = 4 or S = 78 negative result S = S35 experiment failed S = 9S4 experiment not completed S = 190S
(A sample result S35 means that if we add digits 35 from the right hand side to a digit sequence then we shall get the digit sequence corresponding to a failed experiment)
You are to write a program which decrypts given sequences of digits.
Input
A integer n stating the number of encrypted results and then consecutive n lines, each containing a sequence of digits given as ASCII strings.
Output
For each analysed sequence of digits the following lines should be sent to output (in separate lines):
+ for a positive result - for a negative result * for a failed experiment ? for a not completed experiment
In case the analysed string does not determine the experiment result, a first match from the above list should be outputted.
Sample Input
4 78 7835 19078 944
Sample Output
+ - ? *
1 #include<iostream> 2 #include<string> 3 #include<stdlib.h> 4 #include<stdio.h> 5 using namespace std; 6 7 int main(void) 8 { 9 //freopen("621.in", "r", stdin); 10 long long n; 11 scanf("%lld", &n); 12 for (long long i = 0; i < n; i++) 13 { 14 string s; 15 int flag; 16 cin >> s; 17 if (s == "1" || s == "4" || s == "78") { printf("+ "); continue; } 18 else if (s[s.size() - 1] == '5' && s[s.size() - 2] == '3') { printf("- "); continue; } 19 else if (s[0] == '9' && s[s.size() - 1] == '4') { printf("* "); continue; } 20 else if (s[0] == '1' && s[1] == '9' && s[2] == '0') { printf("? "); continue; } 21 22 } 23 //system("pause"); 24 return 0; 25 }