1：
#include <bits/stdc++.h>
using namespace std;
int main(){
for(int i = 300; i <= 500; i += 2){
if(i % 3 == 0 && i % 7 == 0) printf("%d\n", i);
}
return 0;
}
2：
#include <bits/stdc++.h>
using namespace std;
int main(){
string s;
cin >> s;
int l = s.length();
int k = 1, ans = 0;
for(int i = l - 1; i >= 0; i--){
int x;
if(s[i] >= 'A' && s[i] <= 'F'){
x = s[i] - 'A' + 10;
}else{
x = s[i] - '0';
}
ans += x * k;
k *= 16;
}
cout << ans << endl;
return 0;
}
3：
#include <bits/stdc++.h>
using namespace std;
int compare(string s1, string s2){
int l1 = s1.length();
int l2 = s2.length();
int i;
for(i = 0; i < min(l1,l2); i++){
if(s1[i] != s2[i]){
return s1[i] - s2[i];
}
}
//这个我觉着我多虑了，考虑到两字符串不等长
if(l1 > i) return s1[i] - '\n';
else if(l2 > i) return s2[i] - '\n';
return 0;
}
int main(){
string s1, s2;
cin >> s1 >> s2;
cout << compare(s1, s2) << endl;
return 0;
}
4：
#include <bits/stdc++.h>
using namespace std;
int main(){
string x, y;
cin >> x >> y;
//reverse反转字符串，stoi将字符串转换为整型
reverse(x.begin(), x.end());
reverse(y.begin(), y.end());
string sum = to_string(stoi(x) + stoi(y));
reverse(sum.begin(), sum.end());
cout << stoi(sum) << endl;
return 0;
}
5：
#include <bits/stdc++.h>
using namespace std;
//核心：NEXT i = 0表示往右走，i = 1表示往下左走，i = 2表示往上走
//根据DFS，一路走到头再选择换路径，优先 i = 1。。。
int NEXT[3][2] = {{0,1}, {1,-1}, {-1,0}};
int n, a[1005][1005];
//book数组用于检查下标 book[x][y] x,y这个坐标点是否登记过
bool book[1005][1005];
void dfs(int x, int y, int k){
for(int i = 0; i < 3; i++){
int tx = x + NEXT[i][0];
int ty = y + NEXT[i][1];
//注意y的范围!!
if(x < 1 || x > n || y < 1 || y > n - x + 1)continue;
if(!book[tx][ty]){
book[tx][ty] = 1;
a[tx][ty] = k;
dfs(tx, ty, k + 1);
}
}
}
int main() {
int k = 1;
cin >> n;
memset(book, 0, sizeof(book));
a[1][1] = k;
book[1][1] = 1;
dfs(1, 1, k + 1);
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n - i + 1; j++){
cout << a[i][j] << "\t";
}
cout << endl;
}
return 0;
}
6：
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[35][35], b[35][35], c[35][35];
int n, m;
cin >> n >> m;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
cin >> a[i][j];
b[i][j] = a[i][j];
}
}
for(int k = 1; k < m; k++) {
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
int sum = 0;
for(int t = 0; t < n; t++) {
sum = sum + a[i][t] * b[t][j];
}
c[i][j] = sum;
}
}
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
a[i][j] = c[i][j];
}
}
}
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
cout << a[i][j] << " ";
}
cout << endl;
}
return 0;
}
7：
#include <bits/stdc++.h>
using namespace std;
vector<string> s;
bool cmp(string s1, string s2){
//将 MM//DD//YYYY 转换成 YYYYMMDD
string ss1 = "";
ss1 = ss1 + s1[6] + s1[7] + s1[8] + s1[9] + s1[0] + s1[1] + s1[3] + s1[4];
string ss2 = "";
ss2 = ss2 + s2[6] + s2[7] + s2[8] + s2[9] + s2[0] + s2[1] + s2[3] + s2[4];
// 16110505 与 17110505 6<7 则ss1 < ss2 s1 < s2 日期上 s1比s2早 需要交换
for(int i = 0; i < ss1.length(); i++){
if(ss1[i] < ss2[i]) return 1;
else if(ss1[i] > ss2[i]) return 0;
}
return 0;
}
/*
测试数据
4
02/02/2002
01/05/2002
05/06/1987
03/03/1000
*/
int main() {
int n;
cin >> n;
for(int i = 0; i < n; i++){
string x;
cin >> x;
s.push_back(x);
}
for(int i = 0; i < 3; i++){
for(int j = 0; j < n - 1; j++){
if(cmp(s[j], s[j + 1])){
swap(s[j], s[j + 1]);
}
}
}
cout << endl;
for(int i = n - 1, t = 0; i >= 0 && t < 3;i--, t++){
cout << s[i] << endl;
}
return 0;
}
8：
#include <bits/stdc++.h>
using namespace std;
int mystringSum(char *str){
int sum = 0;
string s = str;
for(int i = 0; i < s.length(); i++){
sum += s[i];
}
return sum;
}
int main() {
char s[1005];
cin >> s;
cout << mystringSum(s);
return 0;
}
9：
标签：892.2021,string,int,s2,s1,++,sum From： https://www.cnblogs.com/rongrongrong/p/16943158.html