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1、数列排序 问题描述 给定一个长度为 n 的数列，将这个数列按从小到大的顺序排列。1 #include #include using namespace std; int main(int argc, char *argv) int n, val; vector v; cin n; for (int i = 0; i val; v.push_back(val); sort(v.begin(), v.end(); for (int i = 0; i #include #include #include using namespace std; string strsixteen16 = “0000
2、“, “1000“, “0100“, “1100“, “0010“, “1010“, “0110“, “1110“,“0001“, “1001“, “0101“, “1101“, “0011“, “1011“, “0111“, “1111“; int main(int argc, char *argv) / freopen(“input.txt“, “r“, stdin); int n; cin n; cin.get(); for (int i = 0; i ans; getline(cin, S); /从十六进制最后一位向前每一位转成 4 位的二进制，且二进制串是逆序的 int pos =
3、S.length() - 1; for (; pos = 0; pos-) if (Spos = 0 i-) cout using namespace std; int main(int argc, char *argv) string str; cin str; long long ans = 0, weight = 1; for (int i = str.size() - 1; i = 0; i-) if (stri = 0 int main(int argc, char *argv) int a; char c16 = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B
4、, C, D, E, F; cin a; stack s; do s.push(ca % 16); a /= 16; while (a); while (!s.empty() cout using namespace std; int main(int argc, char *argv) int Arr5, n; cin n; /五位 for (int i = 100; i using namespace std; int main(int argc, char *argv) int a, b; for (int i = 10; i using namespace std; int main(
5、int argc, char *argv) int a, b, c; for (int i = 100; i using namespace std; int main(int argc, char *argv) int Triangle5050; int n; cin n; for (int i = 0; i using namespace std; int main(int argc, char *argv) int Arr1005, n, key; cin n; for (int i = 0; i Arri; cin key; for (int i = 0; i using namesp
6、ace std; int main(int argc, char *argv) int n, max = -100000, min = 100000, sum = 0, val; cin n; for (int i = 0; i val; if (val max) max = val; if (val using namespace std; int main(int argc, char *argv) int m, n, pos = 25; string str = “ZYXWVUTSRQPONMLKJIHGFEDCBABCDEFGHIJKLMNOPQRSTUVWXYZ“; cin n m; for (int i = 0; i 题目分析：典型的回溯法，枚举各种情况。 #include using namespace std; int Bin10; void backtrack(int cur) if (cur = 5) for (int i = 0 ; i using namespace std; int main(int argc, char *argv) int year; cin year; if (0 = year % 4 else cout “no“ endl; return 0;
