..第一次用C++的运算符重载....

很不习惯....调试了几十分钟= =

发现错误很难改..= =..导致代码非常长(基本是重复的)....

郁闷中..

注:本题数据好象不大..我开到2000位就过了..

rogerduck

字符数组才是王道

楼下的可以去参加IOCCC比赛了

( 2006-10-10 20:26:21 )

IOCCC是C语言的

字符数组才是王道

楼下的可以去参加IOCCC比赛了

啊，貌似用冒泡的就可以了

用ANSISTING就OK了，没有必要什么高精度来处理那么麻烦！注意的是：字符串的比较，用字符串来储存数字，比较的时候要注意，先比较字符串长度，长度相等的再比较字符串大小，还有，之前应该先让NAME排序先，这样可以避免出现相同帖子数时候的情况，

应该是l2:=length(b[j])?

出自OIBH Online Judge..

顺手练了基数排序..

数据十分十分小，用int64或qword就行拉

我100哦~！！！

#include <iostream>
#include <iterator>
#include <set>

struct king{
  std::string sorce;
  std::string name;
};
bool operator<(king lhs, king rhs)
{
  if(lhs.sorce.size() < rhs.sorce.size()) return true;
  if(lhs.sorce.size() > rhs.sorce.size()) return false;
  if(lhs.sorce == rhs.sorce) return lhs.name > rhs.name;
  return lhs.sorce < rhs.sorce;
}
std::ostream& operator<<(std::ostream& out, king k)
{
  out << k.name << std::endl;
  return out;
}

int main([[maybe_unused]] int argc, [[maybe_unused]] const char *argv[])
{
  int n;
  std::cin >> n;
  std::set<king> v;
  for (int i = 0; i < n; ++i) {
    std::string s, n;
    std::cin >> n >> s;
    v.insert({s, n});
  }
  std::copy(v.crbegin(), v.crend(),
      std::ostream_iterator<king>(std::cout));
}

评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 588 KiB, score = 5
测试数据 #1: Accepted, time = 0 ms, mem = 596 KiB, score = 15
测试数据 #2: Accepted, time = 0 ms, mem = 592 KiB, score = 20
测试数据 #3: Accepted, time = 0 ms, mem = 592 KiB, score = 25
测试数据 #4: Accepted, time = 0 ms, mem = 592 KiB, score = 35
Accepted, time = 0 ms, mem = 596 KiB, score = 100
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct water {
    string name,num;
}v[1001];
int n;
int cmp(water x,water y) {
    if(x.num.length() != y.num.length()) return x.num.length() > y.num.length();
    if (x.num != y.num) return x.num>y.num;
    return x.name<y.name;
    return 0;
}
void solve() {
    scanf("%d",&n);
    for (int i = 1;i <= n;i++) cin >> v[i].name >> v[i].num;
    sort(v+1,v+n+1,cmp);
    for (int i = 1;i <= n;i++) cout << v[i].name << endl;
}
int main() {
    solve();
    return 0;
}

看那么多神犇都写那么一大串……作为蒟蒻我好慌。
不再多说，思路如下
读入数据
把位数小的高精度高位补零
直接用string的compare。
c++
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
struct info{
string name, num;
friend bool operator <(info a,info b){
if(a.num!=b.num) return a.num>b.num;
return a.name<b.name;
}
};
int main(){
info s[1000];/*
,，
.。
:：
;；
-—
?？ */
int n;
cin>>n;
unsigned int length=0;
for(int i=0;i<n;i++){
cin>>s[i].name>>s[i].num;
if(s[i].num.size()>length) length=s[i].num.size();
}
for(int i=0;i<n;i++){
if(s[i].num.size()<length){
for(int j=s[i].num.size();j<=length;j++){
s[i].num='0'+s[i].num;
}
}
}
sort(s,s+n);
for(int i=0;i<n;i++){
cout<<s[i].name<<endl;
}
return 0;
}


var s1,s2:array[1..1000]of ansistring;
n,x,y,i,j:longint;

procedure change( var x,y:ansistring);
var t:ansistring;
begin
t:=x;x:=y;y:=t;
end;

begin
readln(n);
for i:=1 to n do
begin
readln(s1[i]);
readln(s2[i]);
end;
for i:=1 to n-1 do
for j:=1 to n-i do
begin
x:=length(s2[j]);y:=length(s2[j+1]);
if (x<y)or((x=y)and(s2[j]<s2[j+1]))or((x=y)and(s2[j]=s2[j+1])and(s1[j]>s1[j+1]))then
begin
change(s1[j],s1[j+1]);change(s2[j],s2[j+1]);
end;
end;
for i:=1 to n do writeln(s1[i]);
end.

因为该题的数的长度很大，所以可以用基数排序（也叫“桶排序”）。
另外，要注意**“若几个ID的发贴数相同，则按照ID的字典顺序先后排列。”**
以下为代码
···
c++
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
using namespace std;
const int maxn=1000,maxlen=10000;
struct preson
{
char num[maxlen],name[maxlen];
int len;
}p[maxn],a[10][maxn];
int c[10];
void qsort(int l,int r)
{
int i,j;
struct preson n;
i=l;
j=r;
n=p[(l+r)/2];
while (i<j)
{
while (strcmp(n.name,p[i].name)==1&&i<r) i++;
while (strcmp(n.name,p[j].name)==-1&&j>l) j--;
if (i<=j)
{
swap(p[i],p[j]);
i++;
j--;
}
}
if (i<r) qsort(i,r);
if (j>l) qsort(l,j);
}
int main()
{
int i,j,k,l=0,n;
bool check;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%s",&p[i].name);
scanf("%s",&p[i].num);
p[i].len=strlen(p[i].num);
for(j=0,k=p[i].len-1;j<k;j++,k--) swap(p[i].num[j],p[i].num[k]);
}
do
{
check=false;
memset(c,0,40);
for(i=0;i<n;i++)
if (p[i].len>l)
{
check=true;
a[p[i].num[l]-48][c[p[i].num[l]-48]++]=p[i];
}else a[0][c[0]++]=p[i];
k=0;
for(i=9;i>=0;i--)
for(j=0;j<c[i];j++)
p[k++]=a[i][j];
l++;
}while(check);
for(i=0;i<n;i=j)
{
for(j=i+1;strcmp(p[i].num,p[j].num)==0&&j<n;j++);
if (strcmp(p[i].num,p[j-1].num)==0&&i<j-1) qsort(i,j-1);
}
for(i=0;i<n;i++) printf("%s\n",p[i].name);
return 0;
}
```

若几个ID的发贴数相同，则按照ID的字典顺序先后排列。
简单高精度，不用计算，只是比大小。所以用c++的string。（注意string比较大小是按字母表比大小形式比的，所以string表示数时，不能直接用“>”,“<”,比较，但是可以用“==”；）
例如 "abc">"aaa" "abc">"aaaa" (按照字母表"aaaa"在"abc"前)
所以 "123">"1111" 在string中是正确的，所以用函数，当且仅当字符串长度相等时可以用">","<"。
因为字符串相等时，长度一定相等，所以可以用“==”
***下面pd()的原理！！！ ***

#include <iostream>
#include <string>

using namespace std;

string num[1000];//贴数
string name[1000];//名字

inline bool pd(string a,string b)//num[a]<num[b]```return true;//“数字的string比大小仍然是按字母表形式，所以用函数来比较大小 ”
{
if(a.size()<b.size())return true;
else if(a.size()>b.size())return false;
else if(a<b)return true;
else return false;
}

inline void ksort(int l,int r,bool xiao)//xiao true排贴数，false排字母表
{
if(l==r)return;
string mid;

if(xiao)
mid=num[(l+r)/2];//贴数
else mid=name[(l+r)/2];//字母表

int i=l,j=r;
string temp;
do
{
if(xiao)//贴数
{
while( pd(mid,num[i]) )++i;
while( pd(num[j],mid) )--j;
}else//字母表
{
while( name[i]<mid )++i;
while( name[j]>mid )--j;
}
if(i<=j)
{
temp=num[i];
num[i]=num[j];
num[j]=temp;
temp=name[i];
name[i]=name[j];
name[j]=temp;
++i;
--j;
}
}while(i<=j);
if(xiao)//贴数
{
if(l<j)ksort(l,j,true);
if(i<r)ksort(i,r,true);
}else//字母表
{
if(l<j)ksort(l,j,false);
if(i<r)ksort(i,r,false);
}
return;
}

int main()
{
int n;
int i;
cin>>n;
for(i=0;i<n;++i)
cin>>name[i]>>num[i];//输入
ksort(0,n-1,true);//排序
int len;//挑选贴数相同的拍字母表
for(i=0;i<n;++i)
{
len=0;
for(len=0;len+i+1<n;++len)
{
if( num[len+i+1]!=num[len+i] )break;
}
if(len!=0)ksort(i,i+len,false);//i是起点，len是长度
}
for(i=0;i<n;++i)
cout<<name[i]<<endl;
return 0;
}

只过了第一个点的一般是没看到“若几个ID的发贴数相同，则按照ID的字典顺序先后排列。”这句话。。
~~~
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
struct king{
char name[35];
char num[205];
}w[1005];
int cmp(const void *x,const void *y){
king a=(king)x;
king b=(king)y;
int len1=strlen(a->num),len2=strlen(b->num);
if(len1==len2) {
int x=-strcmp(a->num,b->num);
if(x==0) return strcmp(a->name,b->name);
else return x;
}
else return len2 - len1;
}
int n;
int main(){
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf(" %s",w[i].name);
scanf(" %s",w[i].num);
}
qsort(w,n,sizeof(w[0]),cmp);
for(int i=0;i<n;i++){
printf("%s\n",w[i].name);
}
return 0;
}

评测结果
编译成功

foo.cpp: In function 'bool compare(const char*, const char*)':
foo.cpp:54:1: warning: control reaches end of non-void function [-Wreturn-type]
}
^
测试数据 #0: Accepted, time = 0 ms, mem = 556 KiB, score = 5
测试数据 #1: Accepted, time = 0 ms, mem = 556 KiB, score = 15
测试数据 #2: Accepted, time = 0 ms, mem = 552 KiB, score = 20
测试数据 #3: Accepted, time = 0 ms, mem = 552 KiB, score = 25
测试数据 #4: Accepted, time = 0 ms, mem = 552 KiB, score = 35
Accepted, time = 0 ms, mem = 556 KiB, score = 100

重载运算符
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxM = 10000 + 10;

int N;

class BIGNUMBER {
public:
friend istream& operator >> (istream &, BIGNUMBER &);

BIGNUMBER operator = (const char *num)
{
length = strlen(num);
number = new int [length];
for(int i = 0;i < length; ++i)
number[i] = num[length - i - 1] - '0';
return *this;
}
bool operator < (const BIGNUMBER &X) const {
if(X.length != length) return X.length < length;
for(int i = length - 1; i >= 0; --i)
if(X.number[i] != number[i]) return X.number[i] < number[i];
return 0;
}
bool operator != (const BIGNUMBER &X) const {
if(X.length != length) return 1;
for(int i = length - 1; i >= 0; --i)
if(X.number[i] != number[i]) return 1;
return 0;
}
private:
int *number;
int length;
};

istream& operator >> (istream &is, BIGNUMBER &element) {
char NUMBER[maxM];
is >> NUMBER;
element = NUMBER;
return is;
}

inline bool compare(const char *X, const char *Y) {
int lengthX = strlen(X);
int lengthY = strlen(Y);
int length = min(lengthX, lengthY);
for(int i = 0; i != length; ++i)
if(X[i] != Y[i]) return X[i] < Y[i];
if(lengthX != lengthY) return lengthX < lengthY;
}

struct Data {
char name[25];
BIGNUMBER number;

bool operator < (const Data &X) const {
if(X.number != number) return X.number < number;
return compare(X.name, name);
}
};//people[maxN];

int main() {
scanf("%d", &N);
Data *people = new Data [N + 1];
for(int i = 1; i <= N; ++i) {
scanf("%s", people[i].name);
cin >> people[i].number;
}
sort(people + 1, people + N + 1);
for(int i = N; i != 0; --i) {
cout << people[i].name;
printf("\n");
}
delete [] people;
people = NULL;
return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct water{
string name;
string num;
}f[1001];
int n;
int comp(const water&x,const water&y)
{
if(x.num.length()>y.num.length()){
return 1;
}
if(x.num.length()<y.num.length()){
return 0;
}
if(x.num>y.num){
return 1;
}
if(x.num<y.num){
return 0;
}
if(x.name<y.name){
return 1;
}
return 0;
}
void init()
{
int i,j;
scanf("%d",&n);
for(i=1;i<=n;i++){
cin>>f[i].name>>f[i].num;
}
sort(f+1,f+n+1,comp);
}
void print()
{
int i;
for(i=1;i<n;i++){
cout<<f[i].name<<endl;
}
cout<<f[n].name;
}
int main()
{
init();
print();
return 0;
}

双关键字排序

program cc;
var n,i,j:longint;
b,p:ansistring;
s:array[1..10000]of ansistring;
name:array[1..1000]of ansistring;
begin
readln(n);
for i:=1 to n do
begin
readln(name[i]);
readln(s[i]);
end;
begin
for i:=1 to n-1 do
for j:=i+1 to n do
if (length(s[i])<length(s[j]))or((s[i]=s[j])
and (name[i]>name[j]))or ((length(s[i])=length(s[j]))and (s[i]<s[j]))then
begin
p:=s[i];s[i]:=s[j];s[j]:=p;
b:=name[i];name[i]:=name[j];name[j]:=b;
end;
end;
for i:=1 to n do
writeln(name[i]);
end.

AC50纪念
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AC AC AC AC
AC AC AC ACACAC
AC AC AC AC
AC AC AC AC

AC ACACACAC
AC AC
ACACAC AC
AC