fly away!

自以为对二进制有很深的理解，面对这种水题，切了好久。。。 题目链接 #include #include typedef long long  ll; bool compare(ll a, ll b, ll c)//防止大数相加溢出 { if (a > 0 && b > 0) { if (a > LLONG_MAX - b) return true; } if

#include #include #include #include #include using namespace std; class st { private: vector vec; public: st(){}; ~st(){}; void pop() { if (vec.size() == 0) { printf("%s\n", "Inv

题目链接 int main() { char a[100], b[100]; int a1, b1, c1, a2, b2, c2,a3,b3,c3; scanf("%d.%d.%d", &a1, &b1, &c1); scanf("%d.%d.%d", &a2, &b2, &c2); int ca1, ca2;//表示进位 c3 = (c1 + c2) % 29; ca1 =

点击打开链接 老TLE，可以看出浮点运算的复杂程度远大于整数运算。怎么优化。。。 int main() { float f1, f2; int dignum; scanf("%d", &dignum); scanf("%f %f", &f1, &f2); float t1 = f1, t2 = f2; int count1 = 0, count2 = 0; while (t

counting the pat int main() { int p=0, pa=0, pat = 0; char s[100000] = { 0 }; scanf("%s", s); int i = 0; while (s[i]!=0) { switch (s[i]) { case 'P': p++; break; case 'A': pa += p; pa

本以为暴力枚举会超时，但竟然过了。。。 #include #include #include int longestPalindrome(string s) { int left = 0, right = s.length() - 1; //为奇数时 int len1 = 0; int len2 = 0; int start1 = left;//标记起始点 int sta

题目链接 参考了大神的代码 #include #include class Solution { public: void gen_path(unordered_map>& father, vector& path, string start, string word,vector>& ret) { path.push_back(word); if (start == w

题目链接 有点动态规划的味道。 #include int main() { char s[100]; int p=0, pa=0, pat=0; scanf("%s", s); int len = strlen(s); for (int i = 0; i { switch (s[i]) { case 'p': p += 1; break; case 'a':