ztr loves lucky numbers
题目链接：
http://acm.hust.edu.cn/vjudge/contest/121332#problem/I
Description
ztr loves lucky numbers. Everybody knows that positive integers are lucky if their decimal representation doesn't contain digits other than 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Lucky number is super lucky if it's decimal representation contains equal amount of digits 4 and 7. For example, numbers 47, 7744, 474477 are super lucky and 4, 744, 467 are not.
One day ztr came across a positive integer n. Help him to find the least super lucky number which is not less than n.
Input
There are T cases
For each cases:
The only line contains a positive integer . This number doesn't have leading zeroes.
Output
For each cases
Output the answer
Sample Input
2
4500
47
Sample Output
4747
47
题意：
定义super number:
有且仅有4和7两个数字，且两个数字出现的次数相同.
给出n，求最小的不小于n的super number
题解：
xjb打了一通大模拟，结果越打越觉得坑；
还好分清细节后过掉了；
模拟思路：考虑当前数位能是否能放4或7(用strcmp比较后续数);
当两数大小确定后，后面的序列应按最小顺序编排；
另解：
- 直接打表然后二分.
- 用next_permutation获取可能的组合再比较.
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define eps 1e-8
#define maxn 3100
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;
char num[25];
int cnt;
int main(int argc, char const *argv[])
{
//IN;
//freopen("out.txt","w",stdout);
// printf("%d\n", 1000000);
// for(int i=1; i<=1000000; i++)
// printf("%d\n", i);
int t; cin >> t; getchar();
//int t = 0;
while(t--)
{
LL n=0,cmps=0;
char c;
cnt = 0;
memset(num,0,sizeof(num));
gets(num);
for(int i=0; i<strlen(num); i++) {
if(num[i]<'0' || num[i]>'9') break;
cnt++;
n = n*10 + num[i] - '0';
}
num[cnt] = 0;
if(cnt&1) {
for(int i=1; i<=(cnt+1)/2; i++) printf("4");
for(int i=1; i<=(cnt+1)/2; i++) printf("7");
printf("\n");
continue;
}
for(int i=1; i<=cnt/2; i++) cmps = cmps*10 + 7;
for(int i=1; i<=cnt/2; i++) cmps = cmps*10 + 4;
if(cmps < n) {
for(int i=1; i<=(cnt+2)/2; i++) printf("4");
for(int i=1; i<=(cnt+2)/2; i++) printf("7");
printf("\n");
continue;
}
char ans[25] = {0};
int si=cnt/2,qi=cnt/2;
for(int i=0; i<cnt; i++) {
if(num[i]<'4') {
if(si) ans[i] = '4', si--;
else ans[i] = '7', qi--;
for(int j=0; j<=i; j++) printf("%c", ans[j]);
while(si--) printf("4");
while(qi--) printf("7");
printf("\n");
break;
}
if(num[i] == '4') {
char tmp[25] = {0};
for(int j=0; j<qi; j++) tmp[j] = '7';
for(int j=qi; j<qi+si-1; j++) tmp[j] = '4';
if(!si || strcmp(tmp,num+i+1) < 0) {
ans[i] = '7'; qi--;
for(int j=0; j<=i; j++) printf("%c", ans[j]);
while(si--) printf("4");
while(qi--) printf("7");
printf("\n");
break;
}
if(strcmp(tmp,num+i+1) == 0) {
ans[i] = '4'; si--;
for(int j=0; j<=i; j++) printf("%c", ans[j]);
printf("%s",tmp);
printf("\n");
break;
}
ans[i] = '4'; si--;
continue;
}
if(num[i] == '7') {
char tmp[25] = {0};
for(int j=0; j<qi-1; j++) tmp[j] = '7';
for(int j=qi-1; j<qi+si-1; j++) tmp[j] = '4';
if(!qi || strcmp(tmp,num+i+1) < 0) {
for(int i=1; i<=(cnt+2)/2; i++) printf("4");
for(int i=1; i<=(cnt+2)/2; i++) printf("7");
printf("\n");
break;
}
if(strcmp(tmp,num+i+1) == 0) {
ans[i] = '7'; qi--;
for(int j=0; j<=i; j++) printf("%c", ans[j]);
printf("%s",tmp);
printf("\n");
break;
}
ans[i] = '7'; qi--;
continue;
}
ans[i] = '7'; qi--;
for(int j=0; j<=i; j++)
printf("%c", ans[j]);
while(si--) printf("4");
while(qi--) printf("7");
printf("\n");
break;
}
//printf("\n");
}
return 0;
}