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  • [csu1392]YY一下

    题意:给定x,求有多少个10^8以内的数满足这个数乘以x以后,最高位到了最低位。设最高位的数字和剩余长度,列等式推理即可。

     1 #pragma comment(linker, "/STACK:10240000,10240000")
     2 
     3 #include <iostream>
     4 #include <cstdio>
     5 #include <algorithm>
     6 #include <cstdlib>
     7 #include <cstring>
     8 #include <map>
     9 #include <queue>
    10 #include <deque>
    11 #include <cmath>
    12 #include <vector>
    13 #include <ctime>
    14 #include <cctype>
    15 #include <set>
    16 
    17 using namespace std;
    18 
    19 #define mem0(a) memset(a, 0, sizeof(a))
    20 #define lson l, m, rt << 1
    21 #define rson m + 1, r, rt << 1 | 1
    22 #define define_m int m = (l + r) >> 1
    23 #define Rep(a, b) for(int a = 0; a < b; a++)
    24 #define lowbit(x) ((x) & (-(x)))
    25 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
    26 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
    27 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
    28 
    29 typedef double db;
    30 typedef long long LL;
    31 typedef pair<int, int> pii;
    32 typedef multiset<int> msi;
    33 typedef multiset<int>::iterator msii;
    34 typedef set<int> si;
    35 typedef set<int>::iterator sii;
    36 typedef vector<int> vi;
    37 
    38 const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
    39 const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1};
    40 const int maxn = 1e5 + 7;
    41 const int maxm = 1e5 + 7;
    42 const int maxv = 1e7 + 7;
    43 const int MD = 1e9 +7;
    44 const int INF = 1e9 + 7;
    45 const double PI = acos(-1.0);
    46 const double eps = 1e-10;
    47 
    48 int digit(LL x) {
    49     int cnt = 0;
    50     while (x) {
    51         cnt++;
    52         x /= 10;
    53     }
    54     return cnt;
    55 }
    56 
    57 int main() {
    58     //freopen("in.txt", "r", stdin);
    59     double tx;
    60     while (cin >> tx) {
    61         LL x = (int)(tx * 10000 + 0.5), get = 0;
    62         if (x >= 100000) {
    63             puts("No solution");
    64             continue;
    65         }
    66         LL p = 1;
    67         for (int i = 0; i <= 7; i++) {
    68             for (int k = 1; k <= 9; k++) {
    69                 LL tmp = k * (x * p - 1e4);
    70                 if (tmp % (LL)(1e5 - x)) continue;
    71                 tmp /= 1e5 - x;
    72                 if (digit(tmp) == i) {
    73                     printf("%d", k);
    74                     if (tmp > 0) printf("%d", tmp);
    75                     puts("");
    76                     get = 1;
    77                 }
    78             }
    79             p *= 10;
    80         }
    81         if (!get) puts("No solution");
    82     }
    83     return 0;
    84 }
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  • 原文地址:https://www.cnblogs.com/jklongint/p/4419007.html
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