https://blog.haitai-social.com/note/%E7%B3%BB%E7%BB%9F%E7%90%86%E8%AE%BA%E8%AF%BE%E7%A8%8B/%E7%AE%97%E6%B3%95%E7%AB%9E%E8%B5%9B%E8%BF%9B%E9%98%B6%E6%8C%87%E5%8D%97/Recent content in 算法竞赛进阶指南 on 蛇索尼 AidenHugo -- gohugo.iozh-CNshesl-meow@qq.com (佘崧林)shesl-meow@qq.com (佘崧林)Mon, 28 Feb 2022 01:19:43 +0800https://blog.haitai-social.com/note/%E7%B3%BB%E7%BB%9F%E7%90%86%E8%AE%BA%E8%AF%BE%E7%A8%8B/%E7%AE%97%E6%B3%95%E7%AB%9E%E8%B5%9B%E8%BF%9B%E9%98%B6%E6%8C%87%E5%8D%97/1.%E5%9F%BA%E6%9C%AC%E7%AE%97%E6%B3%95/Mon, 28 Feb 2022 01:19:43 +0800shesl-meow@qq.com (佘崧林)https://blog.haitai-social.com/note/%E7%B3%BB%E7%BB%9F%E7%90%86%E8%AE%BA%E8%AF%BE%E7%A8%8B/%E7%AE%97%E6%B3%95%E7%AB%9E%E8%B5%9B%E8%BF%9B%E9%98%B6%E6%8C%87%E5%8D%97/1.%E5%9F%BA%E6%9C%AC%E7%AE%97%E6%B3%95/位运算 快速幂： 1 2 3 4 5 6 7 8 int powmod(int a, int b, int p) { int ans = 1 % p; for (; b; b >>= 1) { if (b & 1) ans = ((long long)ans * a) % p; a = ((long long)a * a) % p; } return ans; } 大数乘法取模（可能溢