• 洛谷1537(bitset+01背包)


    总数是偶数并且其一半可得即可。

    bitset的移位可替代原本的数组转移。

     1 #pragma comment(linker, "/STACK:1024000000,1024000000")
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cstdlib>
     5 #include <cmath>
     6 #include <ctime>
     7 #include <cctype>
     8 #include <climits>
     9 #include <iostream>
    10 #include <iomanip>
    11 #include <algorithm>
    12 #include <string>
    13 #include <sstream>
    14 #include <stack>
    15 #include <queue>
    16 #include <set>
    17 #include <map>
    18 #include <vector>
    19 #include <list>
    20 #include <fstream>
    21 #include <bitset>
    22 #define init(a, b) memset(a, b, sizeof(a))
    23 #define rep(i, a, b) for (int i = a; i <= b; i++)
    24 #define irep(i, a, b) for (int i = a; i >= b; i--)
    25 using namespace std;
    26 
    27 typedef double db;
    28 typedef long long ll;
    29 typedef unsigned long long ull;
    30 typedef pair<int, int> P;
    31 const int inf = 0x3f3f3f3f;
    32 const ll INF = 1e18;
    33 
    34 template <typename T> void read(T &x) {
    35     x = 0;
    36     int s = 1, c = getchar();
    37     for (; !isdigit(c); c = getchar())
    38         if (c == '-')    s = -1;
    39     for (; isdigit(c); c = getchar())
    40         x = x * 10 + c - 48;
    41     x *= s;
    42 }
    43 
    44 template <typename T> void write(T x) {
    45     if (x < 0)    x = -x, putchar('-');
    46     if (x > 9)    write(x / 10);
    47     putchar(x % 10 + '0');
    48 }
    49 
    50 template <typename T> void writeln(T x) {
    51     write(x);
    52     puts("");
    53 }
    54 
    55 const int maxn = 2e4 + 5;
    56 int a[7], sum, kase;
    57 bitset<maxn> bt;
    58 
    59 int main() {
    60     while (233) {
    61         sum = 0;
    62         rep(i, 1, 6)    read(a[i]), sum += a[i] * i;
    63         if (!sum)    break;
    64         printf("Collection #%d:
    ", ++kase);
    65         if (sum & 1)    puts("Can't be divided.
    ");
    66         else {
    67             bt.reset(); bt.set(0);
    68             rep(i, 1, 6)    rep(j, 1, a[i])    bt |= bt << i;
    69             if (bt.test(sum >> 1))    puts("Can be divided.
    ");
    70             else    puts("Can't be divided.
    ");
    71         }
    72     }
    73     return 0;
    74 }
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  • 原文地址:https://www.cnblogs.com/AlphaWA/p/10534041.html
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