#include<bits/stdc++.h>usingnamespace std;#definea12#defined3intmain(){
int n =0;while(cin >> n){
int res =0;int an = a1 +(n -1)* d;
res =(a1 + an)* n /2;
cout << res << endl;}return0;}
101.输入整型数组和排序标识,对其元素按照升序或降序进行排序
#include<bits/stdc++.h>usingnamespace std;structcmp_0{
booloperator()(constint& a,constint& b){
return a < b;}};boolcmp_1(constint& a,constint& b){
return a > b;}intmain(){
int n =0;while(cin >> n){
vector<int>nums(n,0);for(int i =0; i < n; i++){
int num =0;
cin >> num;
nums[i]= num;}int m =0;
cin >> m;if(m ==0){
//升序sort(nums.begin(), nums.end(),cmp_0());for(int i =0; i < n; i++){
cout << nums[i]<<" ";}
cout << endl;}elseif(m ==1){
//降序sort(nums.begin(), nums.end(), cmp_1);for(int i =0; i < n; i++){
cout << nums[i]<<" ";}
cout << endl;}}}
#include<bits/stdc++.h>usingnamespace std;intmain(){
int n =0;while(cin >> n){
vector<int>nums(n,0);for(int i =0; i < n; i++){
int num =0;
cin >> num;
nums[i]= num;}
vector<int>dp(n,1);//dp[i]表示以nums[i]为结尾的严格上升子序列的长度(初始化为1)int res =1;for(int i =0; i < nums.size(); i++){
for(int j =0; j < i; j++){
//遍历前面的if(nums[j]< nums[i]){
//确保严格上升
dp[i]=max(dp[i], dp[j]+1);//}}
res =max(res, dp[i]);//维护最大值}
cout << res << endl;}return0;}
105.记负均正II
#include<bits/stdc++.h>usingnamespace std;intmain(){
int num =0;double sum =0, ave =0;int posNum =0;int negNum =0;while(cin >> num){
if(num <0){
negNum++;}else{
posNum++;
sum += num;}if(posNum !=0) ave = sum / posNum;}
cout << negNum << endl;printf("%.1lf\n", ave);}
106.字符逆序
#include<bits/stdc++.h>usingnamespace std;voidprocess(string str, string& res){
reverse(str.begin(), str.end());
res = str;}intmain(){
string str ="";while(getline(cin, str)){
string res ="";process(str, res);
cout << res << endl;}return0;}
107.求解立方根
#include<bits/stdc++.h>usingnamespace std;doublecal(double x){
//正负数都有double l =min(-1.0, x);double r =max(1.0, x);double y;while(abs(r - l)>0.01){
//立方根的精度值
y =(l + r)/2;if(y*y*y > x){
//
r = y;}else{
l = y;}}return y;}intmain(){
double num =0.0;while(cin >> num){
//cout << setprecision(1) << fixed << cal(num) << endl; //控制小数位输出printf("%.1lf",cal(num));}return0;}
108.求最小公倍数
#include<bits/stdc++.h>usingnamespace std;intmain(){
int A =0, B =0;while(cin >> A >> B){
int tmp = A * B;for(int i =min(A, B); i <= tmp; i++){
if(i % A ==0&& i % B ==0){
cout << i << endl;break;}}}}