做一题更新一题，题目顺序比较随机，待全部做完再好好整理
/*
1001
A+B Format
if else 语句中，用升序去做有个测试点通不过
#include
using namespace std;
int main()
{
long a,b,c;
cin>>a>>b;
c=a+b;
if(c<0)
{
cout<<"-";
c=-c;
}
if(c>=1000000){
printf("%d,%03d,%03d",c/1000000,c%1000000/1000,c%1000);
}else if(c>=1000){
printf("%d,%03d",c/1000,c%1000);
}else{
printf("%d",c);
}
}
*/
/*
1002
A+B for Polynomials
多项式求和，用数组做 注意输出格式
#include
using namespace std;
int main()
{
float a[1001]={0};
int m,n,max1=0,max2=0,max;
cin>>m;
for(int i=0;i<m;i++)
{
int zhishu;
cin>>zhishu;
float temp;
cin>>temp;
a[zhishu]+=temp;
if(zhishu>max1)max1=zhishu;
}
cin>>n;
for(int j=0;j<n;j++)
{
int zhishu;
cin>>zhishu;
float temp;
cin>>temp;
a[zhishu]+=temp;
if(zhishu>max2)max2=zhishu;
}
max=(max1>max2?max1:max2);
int count=0;
for(int i=0;i<=max;i++)
{
if(a[i]!=0)count++;
}
cout<<count;
for(int i=max;i>=0;i--)
{
if(a[i]!=0)
printf(" %d %.1f",i,a[i]);
}
return 0;
}
*/
/*
1005
Spell It Right 给一个数，计算各位数和，用英文输出
#include
using namespace std;
int main()
{
string s;
cin>>s;
int sum=0;
for(int i=0;i<s.size();i++)
{
sum+=s[i]-'0';
}
string spell[10]={"zero","one","two","three","four","five","six","seven","eight","nine"};
string res = "";
while(sum >= 10){
res = " " + spell[sum%10] + res; //字符串骚操作 一步一步往前走
sum /= 10;
}
if(sum < 10)
res = spell[sum] + res;
cout << res << endl;
return 0;
}
*/
/*
1006
Sign In and Sign Out 输出最早来和最晚走的id
#include
using namespace std;
int main()
{
int n;
cin>>n;
string max;
string min;
int xiao=999999;
int da=000000;
for(int i=0;i<n;i++)
{
string num;
int a,b,c,d,e,f;
int lai,qu;
cin>>num;
scanf("%d:%d:%d %d:%d:%d",&a,&b,&c,&d,&e,&f);
lai=a10000+b100+c;
qu=d10000+e100+f;
if(lai<xiao)
{
min=num;
xiao=lai;
}
if(qu>da){
max=num;
da=qu;
}
}
cout<<min<<" "<<max;
return 0;
}
*/
/*
1008
Elevator
#include
using namespace std;
int main()
{
int count;
cin>>count;
int a[count]={0};
for(int i=0;i<count;i++)
{
cin>>a[i];
}
int sum=0;
int temp=0;
for(int i=0;i<count;i++)
{
if(a[i]>temp)sum+=(a[i]-temp)*6+5;//上升
else sum+=(temp-a[i])*4+5; //下降或原地不动
temp=a[i];
}
cout<<sum;
return 0;
}
*/
/*
1007
Maximum Subsequence Sum 求最大子串和 输出和与开始，结束的值
***思路：遍历开头，再遍历后面的和，有新的大值更新max并更新头尾 ***
#include
using namespace std;
int main()
{
int n;
cin>>n;
int a[n] = {0};
int begin,end;
for(int i = 0; i< n; i++)
{
cin>>a[i];
}
int max = -1;
for(int i = 0; i<n-1; i++){
int temp = 0;
for(int j = i; j< n; j++){
temp += a[j];
if(temp > max){
max = temp;
begin = a[i];
end = a[j];
}
}
}
if(max<0){
max = 0;
begin = a[0];
end = a[n-1];
}
cout<<max<<" "<<begin<<" "<<end<<endl;
return 0;
}
//标准答案
#include
int a[10001];
int main(){
int n, sum = -1, tmp, start, end;
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%d", &a[i]);
for (int i = 0; i < n; i++){
tmp = 0;
for (int j = i; j < n; j++){
tmp += a[j];
if (tmp > sum){
sum = tmp;
start = a[i];
end = a[j];
}
}
}
if (sum < 0){
printf("0 %d %d\n", a[0], a[n-1]);
return 0;
}
printf("%d %d %d\n", sum, start, end);
return 0;
}
*/
/*
1011
World Cup Betting
#include
using namespace std;
int main()
{
char c[3]={'W','T','L'};
double sum=1.0,max=0;
int count;
for(int j=0;j<3;j++)
{
double max=0;
for(int i=0;i<3;i++)
{
double a[3]={0.0};
// scanf("lf",&a[i]);
cin>>a[i];
if(a[i]>max)
{
max=a[i];
count=i;
}
}
cout<<c[count]<<" ";
sum*=max;
}
sum=(sum*0.65-1)*2;
printf("%.2f",sum);
return 0;
}
//标准答案
#include
char S[3] = {'W','T','L'};
int main(){
double ans = 1.0, tmp, a;
int idx; //记录每行最大数字的下标
for(int i = 0; i < 3; i++){
tmp = 0.0;
for(int j = 0; j < 3; j++){ //寻找该行最大的数字存于 tmp
scanf("%lf", &a);
if(a > tmp){
tmp = a;
idx = j;
}
}
ans *= tmp; //按公式累乘
printf("%c ",S[idx]); //输出对应的比赛结果
}
printf("%.2f", (ans * 0.65 - 1) * 2); //输出最大收益
return 0;
}
*/
/*
1025
PAT Ranking 读入区块成绩，分别输出几个排名
思路：node里面填充，第一个参数为小循环次数，小循环里面装入vector，更新local，根据成绩排序，
更新local_rank，放入大的vector，大的vector在小循环执行完后根据成绩更新final_rank
初始化小vector时一开始放在小循环外，然后clear，测试例子通不过，将小vector放入小循环开头初始化
例子可以通过，但是测试点部分通过
通过重新审题，发现问题可能出在cmp函数里，即相同成绩的按序号排序上，更改后AC
#include
#include
#include
using namespace std;
struct node
{
string id;
int score;
int final_rank;
int local;
int local_rank;
};
bool cmp(node a, node b) //按成绩降序
{
return (a.score!=b.score)? a.score>b.score : a.id<b.id;
}
int main()
{
int count_kuai;
vectorvcount;
cin>>count_kuai;
int sum = 0;
for(int i = 0 ; i< count_kuai; i++){
vectorv;
int m;
cin>>m;
sum += m;
for(int j = 0; j<m; j++){
node temp;
cin>>temp.id>>temp.score;
temp.local = i+1; //更新local
v.push_back(temp);
}
//更新local_rank
sort(v.begin(),v.end(),cmp);
int k = 1;
v[0].local_rank = 1;
vcount.push_back(v[0]); for(int j = 1; j<m; j++){
if(v[j].score != v[j-1].score){
k++;
v[j].local_rank = k;
}
else if(v[j].score == v[j-1].score){
v[j].local_rank = v[j-1].local_rank;
k++;
}
vcount.push_back(v[j]);
// v.clear();
// for(int i = m-1; i>=0; i++){
// v.pop_back(v[i]);
// }
}
}
//更新final_rank
sort(vcount.begin(),vcount.end(),cmp);
int t = 1;
vcount[0].final_rank = 1;
for(int i = 1; i<vcount.size(); i++){
if(vcount[i].score != vcount[i-1].score){
t++;
vcount[i].final_rank = t;
}
else if(vcount[i].score == vcount[i-1].score){
vcount[i].final_rank = vcount[i-1].final_rank;
t++;
}
}
//输出
cout<<sum<<endl;
for(int i = 0; i<vcount.size(); i++){
cout<<vcount[i].id<<" "<<vcount[i].final_rank<<" "<<vcount[i].local<<" "<<vcount[i].local_rank<<endl;
}
return 0;
}
*/
/*
1036
Boys vs Girls 给出一系列学生的信息，第一排输出女生中最高成绩对应的名字和学号，
第二排输出男生中最低成绩对应的名字和学号，第三排输出女生与男生成绩的差值
若有对应项没有信息，对应位置输出Absent，第三排输出NA
思路：用两个vector分别存储信息，排序后判断再计算
#include
#include
#include
#include
using namespace std;
struct node
{
string name;
char gender;
string id;
int grade;
};
bool cmp1(node a, node b) //升序
{
return a.grade<b.grade;
}
bool cmp2(node a, node b) //降序
{
return a.grade>b.grade;
}
int main()
{
int n;
cin>>n;
int count_m = 0;
int count_f = 0;
vector m,f;
//分别存入对应向量
for(int i = 0; i<n; i++){ //两种读入方式都可以
// string name;
// char gender;
// string id;
// int grade;
// cin>>name>>gender>>id>>grade;
// m.push_back(node{name,gender,id,grade}); node temp;
cin>>temp.name>>temp.gender>>temp.id>>temp.grade;
if(temp.gender == 'M'){
m.push_back(temp);
count_m++;
}
else if(temp.gender == 'F'){
f.push_back(temp);
count_f++;
}
}
//排序
sort(m.begin(),m.end(),cmp1); //男生升序
sort(f.begin(),f.end(),cmp2);//女生降序
//分类输出
if(count_m>0 && count_f>0){
// printf("%s %s\n",f[0].name.c_str(),f[0].id.c_str());
// printf("%s %s\n",m[0].name.c_str(),m[0].id.c_str());
cout<<f[0].name.c_str()<<" "<<f[0].id.c_str()<<endl;
cout<<m[0].name.c_str()<<" "<<m[0].id.c_str()<<endl;
cout<<f[0].grade-m[0].grade<<endl;
}
else if(count_m == 0 && count_f != 0){
printf("%s %s\n",f[0].name.c_str(),f[0].id.c_str());
cout<<"Absent"<<endl;
cout<<"NA"<<endl;
}
else if(count_f == 0 && count_m != 0){
cout<<"Absent"<<endl;
printf("%s %s\n",m[0].name.c_str(),m[0].id.c_str());
cout<<"NA"<<endl;
}
else if(count_f == 0 && count_m == 0){
cout<<"Absent"<<endl;
cout<<"Absent"<<endl;
cout<<"NA"<<endl;
}
return 0;
}
*/
/*
1037
Magic Coupon 给出两排数据，从中选出相同个数进行计算，输出最大乘积和
思路：用四个vector分别两排的正负数，读取相应个数对应较小的为size为项数 累加求和
#include
#include
#include
using namespace std;
bool cmp1(int a, int b) //升序（对负数用）
{
return a<b;
}
bool cmp2(int a, int b) //降序（对正数用）
{
return a>b;
}
int main()
{
vectorzheng1,zheng2,fu1,fu2;
int n,m;
cin>>n;
for(int i = 0 ; i<n; i++){
int temp;
cin>>temp;
if(temp>0) zheng1.push_back(temp);
else if(temp<0) fu1.push_back(temp);
}
cin>>m;
for(int j = 0; j<m; j++){
int temp;
cin>>temp;
if(temp>0) zheng2.push_back(temp);
else if(temp<0) fu2.push_back(temp);
}
sort(zheng1.begin(),zheng1.end(),cmp2);
sort(zheng2.begin(),zheng2.end(),cmp2);
sort(fu1.begin(),fu1.end(),cmp1);
sort(fu2.begin(),fu2.end(),cmp1);
int size1,size2;
size1 = zheng1.size()<zheng2.size()? zheng1.size() : zheng2.size();
size2 = fu1.size()<fu2.size()? fu1.size() : fu2.size();
int count = 0;
for(int i = 0; i<size1; i++){
count += zheng1[i]*zheng2[i];
}
for(int j = 0; j<size2; j++){
count += fu1[j]*fu2[j];
}
cout<<count<<endl;
return 0;
}
*/
/*
1038
Recover the Smallest Number 给一系列数字，组一个最小数字
思路：字符串数组读入，去判断相同位数上的数字大小，排序
include
using namespace std;
int main()
{ int n;
cin>>n;
string s[n];
for(int i = 0; i<n; i++){
cin>>s[i];
}
//冒泡排序去交换 如何写交换条件就看本事了
for(int i = 0; i<n-1; i++){
for(int j = i+1; j<n; j++){
int k;
for(k = 0;k<(s[i].size()<s[j].size()?s[i].size():s[j].size());k++)
{
string temp;
if(s[i][k]<s[j][k]){
break;
}
else if(s[i][k]>s[j][k]){
temp = s[i];
s[i] = s[j];
s[j] = temp;
break;
}
}
if(k<(s[i].size()<s[j].size()?s[i].size():s[j].size())) continue;
else if(s[i][k] != '\0'){
if(s[i][k]>s[j][0]){
string temp;
temp = s[i];
s[i] = s[j];
s[j] = temp;
// break;
}
}
else if(s[j][k] != '\0'){
if(s[j][k]<s[i][0]){
string temp;
temp = s[i];
s[i] = s[j];
s[j] = temp;
// break;
}
}
}
}
//第一个数去掉左起的零
int t;
for(int j = 0; j<s[0].size(); j++){
if(s[0][j] != '0'){
t = j;
break;
}
}
for(int j = t; j<s[0].size(); j++){
cout<<s[0][j];
}
for(int i = 1; i<n; i++){
cout<<s[i];
}
return 0;
}
//大神思路：比较每两个相邻的元素，如果交换可以使得整个序列变大，就交换之，直到最后没有任何两个值之间能进行交换
注释掉的地方是只考虑第一个字符串左起有零的情况，但是有一个测试点通不过，改为先整合为大字符串再判断左起零，包括全零输出零的情况，AC
#include
#include
using namespace std;
bool cmp(string &a,string &b) //实现大的在前，用地址操作 改变顺序
{
return a+b < b+a;
}
int main()
{
int n;
cin>>n;
string s[n];
for(int i = 0; i<n; i++){
cin>>s[i];
}
sort(s,s+n,cmp);
/*
//第一个数去掉左起的零
int t;
for(int j = 0; j<s[0].size(); j++){
if(s[0][j] != '0'){
t = j;
break;
}
}
for(int j = t; j<s[0].size(); j++){
cout<<s[0][j];
} for(int i = 1; i<n; i++){
cout<<s[i];
}
*/
string out;
for(int i = 0; i<n; i++){
out += s[i];
}
int j;
for( j = 0; j<out.size()&&out[j]=='0'; j++);
if(j==out.size())cout<<"0";
else printf("%s",out.c_str()+j);
return 0;
}
*/
/*
1041
Be Unique 渣渣题
思路：一个数组按顺序存数，另一个数组利用下标存储出现次数
#include
using namespace std;
int main()
{
int a[100001] = {0};
int b[100001]= {0};
int n;
cin>>n;
for(int i = 0 ; i<n; i++){
int temp;
cin>>temp;
a[temp]++;
b[i] = temp;
}
for(int i = 0 ; i<n; i++){
if(a[b[i]] == 1){
cout<<b[i];
return 0;
}
}
cout<<"None"<<endl;
return 0;
}
*/
/*
1042
Shuffling Machine
思路：字符串数组存储，数字数组存储交换顺序
#include
using namespace std;
int main()
{
string a[55] = {"0","S1","S2","S3","S4","S5","S6","S7","S8","S9","S10","S11","S12","S13","H1","H2","H3","H4","H5","H6","H7","H8","H9","H10","H11","H12","H13","C1","C2","C3","C4","C5","C6","C7","C8","C9","C10","C11","C12","C13","D1","D2","D3","D4","D5","D6","D7","D8","D9","D10","D11","D12","D13","J1","J2"};
string b[55] = {"0"};
int c[55] = {0};
int k;
cin>>k;
for(int j = 1; j<55; j++){
int temp;
cin>>temp;
c[j] = temp;
b[temp] = a[j];
}
for(int j = 1; j<55; j++){
a[j] = b[j];
} for(int i = 1; i<k; i++){
for(int j = 1;j<55 ;j++){
b[c[j]] = a[j];
}
for(int j = 1; j<55; j++){
a[j] = b[j];
}
}
cout<<a[1];
for(int j = 2; j<55; j++){
cout<<" "<<a[j];
}
return 0;
}
*/
/*
1044 有两个运行超时 count+min依然超时，神奇
Shopping in Mars 给定n个数和一个应该支付的数，在数列中连续几个刚好够支付或者最小差的支付方式输出
思路：死循环 至flag位，强行遍历输出刚好总和为这个数的方案，若没有，总和+1继续遍历
#include
using namespace std;
int main()
{
int n,count;
scanf("%d %d",&n,&count);
// cin>>n>>count;
int a[n+1] = {0};
for(int i = 1; i<=n; i++){
scanf("%d",&a[i]);
// cin>>a[i];
}
int flag = 0;
int min = 999;
while(1){
for(int j = 1; j<=n; j++){
int sum = 0;
for(int k = 0;k<=n-j;k++){
sum += a[j+k];
if(sum == count){
printf("%d-%d\n",j,j+k);
// cout<<j<<"-"<<j+k<<endl;
flag = 1;
break;
}
if(sum > count){
int temp = sum-count;
if(temp<min)
min = temp;
break;
}
}
}
if(flag) return 0;
else count += min;
}
return 0;
}
*/
/*
1149
Dangerous Goods Packaging
#include
#include
#include
using namespace std;
bool judge(vector<set >v,int *a,int k)
{
for(int i = 0; i<k-1; i++){
for(int j = i+1; j<k; j++){
if(i!=j && v[a[i]].find(a[j]) != v[a[i]].end())
return false;
}
}
return true;
}
int main()
{
int m,n;
cin>>m>>n;
vector<set >v(100010);
for(int i = 0; i<m; i++){
int a,b;
cin>>a>>b;
v[a].insert(b);
v[b].insert(a);
}
for(int j = 0; j<n; j++){
int k;
cin>>k;
int a[k];
for(int i = 0; i<k; i++){
cin>>a[i];
}
printf("%s\n",judge(v,a,k)? "Yes" : "No");
}
return 0;
}
*/
/* 1015 Reversible Primes 给一个整数，给一个进制，判断这个数和进制转换后的相反数(同一进制)是否都为素数，若是则输出Yes 否则输出No
思路：进制转换，判断素数的结合
#include
#include
using namespace std;
bool IsLeap(int n)
{
if(n<2) return false;
for(int i = 2; i<= sqrt(n); i++){
if(n%i == 0)return false;
}
return true;
}
int main()
{
int a[12] = {0};
int b[12] = {0};
while(1){
int count,m;
cin>>count;
int hh = count;
if(count<0){
return 0;
}
else cin>>m;
int k = 0;
while(count>0){
a[k] = count % m;
count = count / m;
k++;
}
int t = 0;
for(int j = k-1; j >= 0; j--){
b[t] = a[j];
t++;
}
int sum = 0;
for(int i = 0; i < k; i++){
sum += b[i]*pow(m,i);
}
if(IsLeap(hh) && IsLeap(sum)) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
}
*/
/*
1027 Colors in Mars 十进制转13进制0-9,ABC，不足两位左起放0，开始有个"#"
#include
using namespace std;
int main()
{
int k = 0;
for(int i = 0 ; i < 3; i++){
char c[14] = {"0123456789ABC"};
int temp;
cin>>temp;
if(k == 0){
cout<<"#";
k++;
}
int high,low;
high = temp / 13;
low = temp % 13;
cout<<c[high]<<c[low];
}
return 0;
}
*/
/*
1031 Hello World for U 给定一个字符串，输出为u型
#include
#include
using namespace std;
int main()
{
string s;
cin>>s;
int len = s.size();
len = len +2;
int lie,hang;
hang = len / 3; //总行数
lie = len - 2*hang; //总列数
for(int i = 0; i < hang-1; i++){
cout<<s[i];
for(int j = 0; j < lie-2; j++)cout<<" ";
cout<<s[s.size()-1-i]<<endl;
}
for(int i = hang-1;i<hang+lie-1;i++)cout<<s[i];
return 0;
}
*/
/*
1035 Password 敏感字符替换
#include
#include
#include
using namespace std;
struct node
{
string id;
string password;
};
int main()
{
int n;
cin>>n;
string id;
string password;
int sum = 0;
vector v;
for(int i = 0 ;i < n; i++){
int flag = 0;
cin>>id>>password;
for(int i = 0 ; i < password.size(); i++){
if(password[i] == '1' || password[i] == '0' || password[i] == 'l' || password[i] == 'O'){
if(password[i] == '1') password[i] = '@';
else if(password[i] == '0') password[i] = '%';
else if(password[i] == 'l') password[i] = 'L';
else if(password[i] == 'O') password[i] = 'o';
flag = 1;
}
}
if(flag == 1){
sum++;
v.push_back(node{id,password});
}
}
if(sum == 0 && n ==1)printf("There is 1 account and no account is modified\n");
else if(sum == 0 && n>1)printf("There are %d accounts and no account is modified\n",n);
else{
cout<<v.size()<<endl;
for(int i = 0;i<v.size();i++){
cout<<v[i].id<<" "<<v[i].password<<endl;
}
}
return 0;
}
*/
/*
1148 Werewolf - Simple Version
#include
#include
using namespace std;
int main()
{
int n;
cin>>n;
int a[n+1];
for(int i = 1; i <= n; i++){
cin>>a[i];
}
for(int i = 1 ; i < n ; i++){
for(int j = i+1 ; j <= n ; j++){
int b[n+1] = {0};
int count_lang = 0;
int count_pm = 0;
b[i] = 1;
b[j] = 1;
for(int k = 1 ; k <= n ; k++){
if(a[k]<0 && (b[abs(a[k])]!=1)){
if(k == i || k == j) count_lang++;
else count_pm++;
}
else if(a[k]>0 && (b[abs(a[k])]!=0)){
if(k == i || k == j) count_lang++;
else count_pm++;
}
}
if(count_lang == 1 &&count_pm == 1){
cout<<i<<" "<<j<<endl;
return 0;
}
}
}
cout<<"No Solution";
return 0;
}
*/
/*
1144 The Missing Number
#include
#include
using namespace std;
int main()
{
int n;
cin>>n;
int a[100010] = {0};
for(int i = 0 ; i < n ; i++){
int k;
cin>>k;
if(k>0 && k<100010)
a[k]++;
}
for(int i = 1 ; i < 100010 ; i++){
if(a[i] == 0){
cout<<i<<endl;
break;
}
}
return 0;
}
*/
/*
1136 A Delayed Palindrome 直接字符串读入 cin>>str 字符串内直接交换
1079 延迟的回文数 多函数操作
#include
using namespace std;
int is(string str)
{
int flag=1;
for(int i=0;i<str.length()/2;i++)
{
if(str[i]!=str[str.length()-i-1])
{
flag=0;
break;
}
}
return flag;
}
int main()
{
int count=0;
string str;
cin>>str;
if(is(str))
{
cout<<str<<" is a palindromic number.";
return 0;
}
while(1)
{
if(count==10)
{
cout<<"Not found in 10 iterations.";
break;
}
string str1,str2;
str1=str2=str;
for(int i=0;i<str1.length()/2;i++)
{
char k=str1[i];
str1[i]=str1[str1.length()-i-1];
str1[str1.length()-i-1]=k; }
int c=0;
for(int i=0;i<str.length();i++)
{
c=c+str[i]-'0'+str1[i]-'0';
str2[i]=c%10+'0';
c/=10; //有进位自动取1，无进位就是0
}
if(c!=0)
{
str2+=c+'0';
}
for(int i=0;i<str2.length()/2;i++)
{
char k=str2[i];
str2[i]=str2[str2.length()-i-1];
str2[str2.length()-i-1]=k; }
cout<<str<<" + "<<str1<<" = "<<str2<<endl;
if(is(str2))
{
cout<<str2<<" is a palindromic number.";
break;
}
str=str2;
count++;
}
return 0;
}
*/
/*
1132 Cut Integer
考虑分母不为零的情况 //提示浮点数错误
#include
#include<string.h>
using namespace std;
int main()
{
int n;
cin>>n;
for(int i = 0 ; i < n ; i++){
string s;
cin>>s;
int sum1 = 0,sum2 = 0,sum3 = 0; //sum = sum10 + 这个数
for(int i = 0 ; i < s.size() ; i++){
sum1 = sum1 * 10 + (s[i] - '0');
}
//前半部分截取
for(int j = 0 ; j < s.size()/2 ; j++ ){
sum2 = sum2 * 10 + (s[j] - '0');
}
//后半部分截取
for(int q = s.size()/2 ; q < s.size() ; q++ ){
sum3 = sum3 * 10 + (s[q] - '0');
}
if(sum2 == 0 || sum3 == 0) cout<<"No"<<endl;
else if(sum1%(sum2sum3) == 0)cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
return 0;
}
*/