HDU 4403 A very hard Aoshu problem

最新推荐文章于 2020-07-12 22:01:39 发布

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本文介绍了一种使用深度优先搜索（DFS）算法解决数串中添加加号以形成等式的暴力求解方法。通过枚举加号位置并判断等式左右两边是否相等来寻找所有可能的解。

dfs.给出一个数串，可以向其中加入若干或0个加号，然后再加一个等号，使等号两端相等。

用dfs枚举加号的位置，然后处理等号两端，判断是否相等，十分暴力也可以过。。。注意用long long，15位数不剪枝还是会超int的

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
///LOOP
#define REP(i, n) for(int i = 0; i < n; i++)
#define FF(i, a, b) for(int i = a; i < b; i++)
#define FFF(i, a, b) for(int i = a; i <= b; i++)
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)
#define FDD(i, a, b) for(int i = a; i >= b; i--)
///INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RFI(n) scanf("%lf", &n)
#define RFII(n, m) scanf("%lf%lf", &n, &m)
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
#define RS(s) scanf("%s", s)
///OUTPUT
#define PN printf("\n")
#define PI(n) printf("%d\n", n)
#define PIS(n) printf("%d ", n)
#define PS(s) printf("%s\n", s)
#define PSS(s) printf("%s ", n)
#define PC(n) printf("Case %d: ", n)
///OTHER
#define PB(x) push_back(x)
#define CLR(a, b) memset(a, b, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(b))
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = 9901;
const int INFI = 1e9 * 2;
const LL LINFI = 1e17;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int N = 22;
const int M = 22;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};

char s[N];
bool add[N];
int l, ans;

LL calc(int a, int b)
{
    LL sum, t;
    sum = t = 0;
    FF(i, a, b)
    {
        t = s[i] - '0';
        while(!add[i] && i < b - 1)t = t * 10 + s[++i] - '0';
        sum += t;
    }
    return sum;
}

bool check(int x)
{
    LL a = calc(0, x + 1);
    LL b = calc(x + 1, l);
    return a == b;
}

void dfs(int x)
{
    if(x == l - 1)
    {
        REP(i, x)if(!add[i] && check(i))ans++;
        return;
    }
    add[x] = 1;
    dfs(x + 1);
    add[x] = 0;
    dfs(x + 1);
}

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

    while(gets(s), s[0] != 'E')
    {
        l = strlen(s);
        CLR(add, 0);
        ans = 0;
        dfs(0);
        PI(ans);
    }
    return 0;
}