POJ-1416 Shredding Company 枚举

给定两个数，这两个数的长度都不超过6位数，然后求后一个数被怎样的分割能使得其和值最接近前一个数，其实这个问题就相当于在后一个数中进行插板而已。通过状态压缩来枚举隔板即可。

代码如下：

#include <iostream>

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <string>

#include <map>

#include <queue>

#include <vector>

#include <stack>

#include <list>

#include <set>

using namespace std;



int t, num, bit[10], idx;



void Break() {

    idx = -1;

    while (num) {

        bit[++idx] = num % 10;

        num /= 10;

    }

    for (int i = 0, j = idx; i < j; ++i, --j) {

        swap(bit[i], bit[j]);

    }

}



int get(int a, int b) {

    int ret = 0;

    for (int i = a; i < b; ++i) {

        ret *= 10, ret += bit[i];

    }

    return ret;

}



void Print(int state) {

    int last = 0;

    for (int i = 1; i <= idx+1; ++i) {

        if (state & (1 << i)) {

            printf(" %d", get(last, i));

            last = i;

        }

    }

}



bool judge(int state, int &ret) {

    int last = 0;

    ret = 0;

    for (int i = 1; i <= idx+1; ++i) {

        if (state & (1 << i)) {

            ret += get(last, i);

            last = i;

        }

    }

    return ret <= t;

} 



int main()

{

    int mask, Max, cnt, state;

    while (scanf("%d %d", &t, &num), t | num) {

        int i;

        if (t == num) {

            printf("%d %d\n", t, num);

            continue;

        }

        mask = cnt = 0;

        Max = 0;

        Break();

        for (i = 1; i <= idx; ++i) {

            mask |= (1 << i); // 对掩码进行赋值

        } 

        for (i = 0; i <= mask; i+=2) {

            int t = i;

            t |= (1 << (idx+1));

            int ret;

            if (judge(t, ret)) {

                if (Max < ret) {

                    Max = ret;

                    state = t;

                    cnt = 1;

                }

                else if (Max == ret) ++cnt;

            }

        }

        if (Max == 0) puts("error");

        else if (cnt != 1) puts("rejected");

        else {

            printf("%d", Max);

            Print(state);

            puts("");

        }

    }

    return 0;

}