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;

}

巴特西

HDU 5676 ztr loves lucky numbers （模拟）