POJ 3340 & HDU 2410 Barbara Bennett's Wild Numbers(数学)
题目链接:
PKU:http://poj.org/problem?id=3340
HDU:http://acm.hdu.edu.cn/showproblem.php?pid=2410
Description
A wild number is a string containing digits and question marks (like 36?1?8). A number X matches a wild number W if they have the same length, and every non-question mark character in X is equal to the character in the same position in W (it means that you can replace a question mark with any digit). For example, 365198 matches the wild number 36?1?8, but 360199, 361028, or 36128 does not. Write a program that reads a wild number W and a number X from input, both of length n, and determines the number of n-digit numbers that match W and are greater than X.
Input
There are multiple test cases in the input. Each test case consists of two lines of the same length. The first line contains a wild number W, and the second line contains an integer number X. The length of input lines is between 1 and 10 characters. The last line of input contains a single character #.
Output
For each test case, write a single line containing the number of n-digit numbers matching W and greater than X (n is the length of W and X).
Sample Input
36?1?8 236428 8?3 910 ? 5 #
Sample Output
100 0 4
Source
题意:
给出两个数字字符串,串一中有‘?‘,问在‘?’位置填数字共有多少中填法,保证串一大于串二!
代码如下:
#include <cstdio> #include <cstring> #include <cmath> const int maxn = 17; typedef __int64 LL; int cont; LL ans; char s1[maxn]; char s2[maxn]; LL POW(LL n, LL k) { LL s = 1; for(LL i = 0; i < k; i++) { s*=n; } return s; } void solve(int len, int k) { for(int i = 0; i < len; i++) { if(s1[i] == '?') { ans+=(9-(s2[i]-'0'))*POW(10,--cont); } else if(s1[i] < s2[i]) return ; else if(s1[i] > s2[i]) { ans+=POW(10,cont); return ; } } } int main() { while(~scanf("%s",s1)) { ans = 0; cont = 0; if(s1[0] == '#') break; scanf("%s",s2); int len = strlen(s1); for(int i = 0; i < len; i++) { if(s1[i] == '?') cont++; } solve(len,0); printf("%I64d\n",ans); } return 0; }
郑重声明:本站内容如果来自互联网及其他传播媒体,其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享,并不代表本站赞同其观点和对其真实性负责,也不构成任何其他建议。