题目大意：引用Worlf of Seven的描述，
What the problem wants is:
3 1 2
32.0 54.7
-2
54.7 is
the 1st position in array
-2 is the 2nd position in array
32.0 is on the
3rd position in array
直接进行转化就好了，还有就是题目要求浮点数的输出和输入形式一样，所以浮点数用字符串存储就好了。
1 #include <cstdio> 2 #include <cctype> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 #define MAXN 1000000 7 8 struct Num 9 { 10 char s[20]; 11 int rank; 12 bool operator < (const Num& n) const 13 { 14 return rank < n.rank; 15 } 16 }; 17 Num num[MAXN]; 18 int idx[MAXN]; 19 20 int main() 21 { 22 #ifdef LOCAL 23 freopen("in", "r", stdin); 24 #endif 25 int T; 26 scanf("%d", &T); 27 getchar(); 28 char str[MAXN]; 29 while (T--) 30 { 31 gets(str); 32 gets(str); 33 int len = strlen(str); 34 int n = 0; 35 for (int i = 0; i < len; ) 36 { 37 if (isdigit(str[i])) 38 { 39 int t = str[i] - '0'; 40 i++; 41 while (isdigit(str[i])) 42 { 43 t = t*10 + str[i] - '0'; 44 i++; 45 } 46 n++; 47 num[n].rank = t; 48 } 49 else i++; 50 } 51 for (int i = 1; i <= n; i++) 52 scanf("%s", num[i].s); 53 sort(num+1, num+n+1); 54 for (int i = 1; i <= n; i++) 55 printf("%s ", num[i].s); 56 getchar(); 57 if (T) printf(" "); 58 } 59 return 0; 60 }
刚开始WA了，看代码感觉没错，看论坛说因为题目没有规定数据大小，所以...于是把MAXN从104改成106，然后就AC了...写程序的时候还感觉104已经够大了呢，看来以后对这种没规定数据量大小的题还是用string，vector这种动态大小的类型好了。
1 #include <cstdio> 2 #include <iostream> 3 #include <cctype> 4 #include <cstring> 5 #include <vector> 6 #include <string> 7 #include <algorithm> 8 using namespace std; 9 10 struct Num 11 { 12 string s; 13 int rank; 14 bool operator < (const Num& n) const 15 { 16 return rank < n.rank; 17 } 18 }; 19 vector<Num> num; 20 21 int main() 22 { 23 #ifdef LOCAL 24 freopen("in", "r", stdin); 25 #endif 26 int T; 27 scanf("%d", &T); 28 getchar(); 29 string str; 30 while (T--) 31 { 32 getline(cin, str); 33 getline(cin, str); 34 int len = str.size(); 35 num.clear(); 36 for (int i = 0; i < len; ) 37 { 38 if (isdigit(str[i])) 39 { 40 int t = str[i] - '0'; 41 i++; 42 while (isdigit(str[i])) 43 { 44 t = t*10 + str[i] - '0'; 45 i++; 46 } 47 num.push_back((Num){"", t}); 48 } 49 else i++; 50 } 51 for (int i = 0; i < num.size(); i++) 52 cin >> num[i].s; 53 sort(num.begin(), num.end()); 54 for (int i = 0; i < num.size(); i++) 55 cout << num[i].s << endl; 56 getchar(); 57 if (T) printf(" "); 58 } 59 return 0; 60 }