(贪心)Codeforces Round #402 B. Weird Rounding
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题目网址: Codeforces Round #402 B. Weird Rounding
题意分析:
题意: 判断一个数是否能被 10^k整除, 如果不能, 求最少次数的去掉某一位数使得这个数能被10^k整除
思路: 从最低位开始计数0的个数, 直到计数到数的最高位或者满足被10^k次方整除为止, 其中 ans 计数非0的个数, 若计数到最高位都不能被 10^k整除, 则结果为这个数的长度-1
代码:
#include <iostream>using namespace std;int main(int argc, char const *argv[]){ int n, k; while (~scanf("%d %d", &n, &k)) { if(n) { int ans = 0; int cnt = 0; while (n && cnt != k) { if(n % 10 != 0) ++ans; else ++cnt; n /= 10; } if(cnt == k) printf("%d\n", ans); else { printf("%d\n", ans+cnt-1); } } else printf("0\n"); } return 0;}
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