hdu 5676-ztr loves lucky numbers

ztr loves lucky numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1022 Accepted Submission(s): 429



Problem 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(1≤n≤105) cases

For each cases:

The only line contains a positive integer n(1≤n≤1018).
This number doesn't have leading zeroes.

Output

For each cases

Output the answer

Sample Input

2
4500
47

Sample Output

4747
47

Source

BestCoder Round #82 (div.2)

解题报告：

纯模拟，自己纸上画一画就知道模拟过程了。注意10的18次方，极限情况下64位会溢出的，所以需要特殊打表处理。另一种思路是打表的方式，所有可能的情况最多是2^10次方，将这些数放进有序数组，然后根据输入进行bound_find查找。代码为前一种解法，有点乱！

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <set>
#include <queue>
#include <vector>
using namespace std;

#define max(x, y) ((x) > (y) ? (x) : (y))
#define min(x, y) ((x) < (y) ? (x) : (y))
typedef long long LL;
LL n;

int GetDigit(LL x)
{
int res = 0;
while (x)
{
++res;
x /= 10;
}
return res;
}

LL GetMaxnumThisDigit(int dig)
{
LL res = 0;
if (dig & 1)
{
return res;
}

for (int i = 0; i < dig/2; ++i)
{
res = res * 10 + 7;
}
for (int i = 0; i < dig/2; ++i)
{
res = res * 10 + 4;
}

return res;
}

LL GetMinnumThisDigit(int dig)
{
LL res = 0;
if (dig & 1)
{
return res;
}

for (int i = 0; i < dig/2; ++i)
{
res = res * 10 + 4;
}
for (int i = 0; i < dig/2; ++i)
{
res = res * 10 + 7;
}

return res;
}

void PrintMinThisDigit(int dig)
{
for (int i = 0; i < dig/2; ++i)
{
printf("4");
}
for (int i = 0; i < dig/2; ++i)
{
printf("7");
}
printf("\n");
}

void Solve()
{
LL ans = 0;

int dig = GetDigit(n);
if ((dig & 1) == 0)
{
// 偶数位
if (n == 0)
{
printf("47\n");
return;
}

if (n > GetMaxnumThisDigit(dig))
{
PrintMinThisDigit(dig+2);
}
else
{
int bits[20] = {0};
int abits[20] = {0};
int tmpId = dig - 1;
LL tmpN = n;
while (tmpN)
{
bits[tmpId--] = tmpN % 10;
tmpN /= 10;
}

int c4 = 0, c7 = 0;
for (int i = 0; i < dig; ++i)
{
if (bits[i] < 4)
{
for (c4; c4 < dig/2; ++c4)
abits[i++] = 4;
for (c7; c7 < dig/2; ++c7)
abits[i++] = 7;
break;
}
else if (bits[i] == 4)
{
if (c4 < dig/2)
{
abits[i] = 4;
++c4;
}
else
{
abits[i] = 7;
++c7;
for (c4; c4 < dig/2; ++c4)
abits[i++] = 4;
for (c7; c7 < dig/2; ++c7)
abits[i++] = 7;
break;
}
}
else if (4 < bits[i] && bits[i] < 7)
{
abits[i++] = 7;
++c7;
for (c4; c4 < dig/2; ++c4)
abits[i++] = 4;
for (c7; c7 < dig/2; ++c7)
abits[i++] = 7;
break;
}
else if (7 == bits[i] && c7 < dig/2)
{
abits[i] = 7;
++c7;
}
else
{
// 向前追溯
for (int j = i - 1; j >= 0; --j)
{
if (abits[j] == 4)
{
abits[j] = 7;
--c4;
++c7;
// 向后填充
for (c4; c4 < dig/2; ++c4)
abits[++j] = 4;
for (c7; c7 < dig/2; ++c7)
abits[++j] = 7;
break;

}
else if (abits[j] == 7)
{
--c7;
}
}
break;
}
}
// calculate
for (int k = 0; k < dig; ++k)
{
printf("%d", abits[k]);
}
printf("\n");
}
}
else
{
// 奇数位
PrintMinThisDigit(dig+1);
}

}

int main()
{
//freopen("input.txt", "r", stdin);

int times;
scanf("%d", ×);
for (int i = 1; i <= times; ++i)
{
//printf("Case #%d:\n", i);

scanf("%lld", &n);
Solve();
}
return 0;
}