CodeForces 449D Jzzhu and Numbers
来源:互联网 发布:mac和nars口红哪个便宜 编辑:程序博客网 时间:2024/05/18 03:02
CodeForces 449D Jzzhu and Numbers
题目描述:
输入
题解:
利用容斥原理,令
考虑所有的二进制下含有
题目链接: vjudge 原网站
代码:
#include <cstdio>#include <cstdlib>#include <algorithm>using namespace std;#define MAXN 1000007#define MAXM 25const long long MOD = 1000000007LL;static int N;static long long ans = 0, bin[MAXN] = {1}, dp[MAXN];int main(){ for (int i = 1; i < MAXN; i++) bin[i] = bin[i-1] * 2 % MOD; scanf("%d", &N); for (int i = 0, x; i < N; i++) scanf("%d", &x), dp[x]++; for (int j = 0; j < MAXM; j++) for (int i = 0; i < MAXN; i++) if ((1 << j) & i) (dp[i ^ (1 << j)] += dp[i]) %= MOD; for (int i = 0; i < MAXN; i++) { int sign = 1; for (int j = 0; j < MAXM; j++) if ((1 << j) & i) sign = -sign; (ans += sign * (bin[dp[i]] - 1) % MOD + MOD) %= MOD; } printf("%lld\n", ans); return 0;}
提交记录(AC / Total = 1 / 1):
0 0
- CodeForces 449D Jzzhu and Numbers
- 449 D. Jzzhu and Numbers
- codeforces 449D Jzzhu and Numbers 容斥+DP
- CodeForces 449D Jzzhu and Numbers 【DP+容斥】
- codeforces 449 D Jzzhu and Numbers(容斥+dp)
- Codeforces Round #257 (Div. 1) D. Jzzhu and Numbers
- CodeForces 449 D.Jzzhu and Numbers(状压DP+容斥原理)
- Codeforces 449B - Jzzhu and Cities / 450D - Jzzhu and Cities
- 【杂题】 codeforces 449A Jzzhu and Chocolate
- Codeforces 449 B. Jzzhu and Cities
- CodeForces 449-A. Jzzhu and Chocolate
- CodeForces 449A Jzzhu and Chocolate
- Codeforces 449 A Jzzhu and Chocolate
- CodeForces 449A - Jzzhu and Chocolate
- Codeforces 449A Jzzhu and Chocolate(贪心)
- Codeforces 449C Jzzhu and Apples(构造)
- CodeForces 449 B. Jzzhu and Cities
- CodeForces 449A - Jzzhu and Chocolate(贪心)
- SQL中登录名的创建
- java 键盘输入三个整数加入到集合中(加入的数不能跟集合中的元素重复)
- python shell快捷键集锦
- 打包
- 析构函数中调用析构函数的情况
- CodeForces 449D Jzzhu and Numbers
- aop注解配置、表达式
- 如何打造一个小而精的电商网站架构?
- 技术积累20170415(4)
- PHP排序算法系列:冒泡排序
- 连表查询
- PHP各种环境下的代码调试
- Android UI性能优化 检测应用中的UI卡顿
- 1028. 人口普查(20)