Countdown

来源：互联网发布：网络监控机编辑：程序博客网时间：2024/05/18 02:12
Countdown

这里写图片描述

.
.
题意：给出6个数，通过+、-、*、/、四种操作，每个数只能用一次，问怎么算能得到一个尽量接近target的数（target给定），然后按题目要求输出。
.
.
解法：这题直接暴力死做不会超时，用map记录数字使用情况以及得出来的值，保存一下是从那两个状态转移过来就好了。.
.
.
#include <iostream>#include <stdio.h>#include <stdlib.h>#include <map>#include <algorithm>#include <vector>using namespace std;struct Node {    int val, state;    Node() {}    Node(int _v, int _s):val(_v), state(_s) {}    void operator=(Node &other) {        val = other.val;        state = other.state;    }    bool operator<(const Node &other) const {        if (val < other.val) return true;        if (val > other.val) return false;        if (state < other.state) return true;        return false;    }    bool operator==(const Node &other) const {        if (val == other.val && state == other.state) return true;        return false;    }    bool operator!=(const Node &other) const {        if (*this == other) return false;        return true;    }};struct Pre {    Node p1, p2;    };map<Node, Pre> store;vector<Node> f[1<<6];Pre d[400000];int val[400000];Node ans;int tot, x, y, tar, a[10], sum, n, m;int main() {    freopen("j.in","r",stdin);    int tt;    Node start(0, 0);    scanf("%d", &tt);    for (int cases = 1; cases <= tt; cases++) {        if (cases > 1) printf("\n");        store.clear();        for (int i = 0; i < 1<<6; i++) f[i].clear();        for (int i = 1; i <= 6; i++) scanf("%d", &a[i]);        scanf("%d", &tar);        ans.val = 1000000;        for (int i = 1; i <= 6; i++) {            Node temp(a[i], 1<<(i-1));            f[1<<(i-1)].push_back(temp);            store[temp].p1 = start;            store[temp].p2 = start;            if (abs(tar-a[i]) < abs(tar-ans.val)) {                ans.val = a[i];                ans.state = 1<<(i-1);            }        }        for (int i = 1; i <= (1<<6)-1; i++) {            for (int jj = 1; jj <= (1<<6)-1;  jj++) {                for (int kk = jj+1; kk <= (1<<6)-1; kk++)                    if (((jj|kk) == i) && ((jj&kk) == 0)) {                    n = f[jj].size();                    m = f[kk].size();                    for (int j = 0; j < n; j++)                        for (int k = 0; k < m; k++) {                            Node temp;                            temp.state = i;                            //  +                            temp.val = f[jj][j].val+f[kk][k].val;                            if (!store.count(temp)) {                                f[i].push_back(temp);                                if (abs(tar-temp.val) < abs(tar-ans.val)) ans = temp;                                store[temp].p1 = f[jj][j];                                store[temp].p2 = f[kk][k];                            }                            //  *                            temp.val = f[jj][j].val*f[kk][k].val;                            if (!store.count(temp)) {                                f[i].push_back(temp);                                if (abs(tar-temp.val) < abs(tar-ans.val)) ans = temp;                                store[temp].p1 = f[jj][j];                                store[temp].p2 = f[kk][k];                            }                            //  -                            temp.val = f[jj][j].val-f[kk][k].val;                            if (temp.val > 0 && !store.count(temp)) {                                f[i].push_back(temp);                                if (abs(tar-temp.val) < abs(tar-ans.val)) ans = temp;                                store[temp].p1 = f[jj][j];                                store[temp].p2 = f[kk][k];                            }                            temp.val = -f[jj][j].val+f[kk][k].val;                            if (temp.val > 0 && !store.count(temp)) {                                f[i].push_back(temp);                                if (abs(tar-temp.val) < abs(tar-ans.val)) ans = temp;                                store[temp].p1 = f[jj][j];                                store[temp].p2 = f[kk][k];                            }                            //  /                            if (f[kk][k].val != 0 && f[jj][j].val%f[kk][k].val == 0) {                                temp.val = f[jj][j].val/f[kk][k].val;                                if (!store.count(temp)) {                                    f[i].push_back(temp);                                    if (abs(tar-temp.val) < abs(tar-ans.val)) ans = temp;                                    store[temp].p1 = f[jj][j];                                    store[temp].p2 = f[kk][k];                                }                            }                            if (f[jj][j].val != 0 && f[kk][k].val%f[jj][j].val == 0) {                                temp.val = f[kk][k].val/f[jj][j].val;                                if (!store.count(temp)) {                                    f[i].push_back(temp);                                    if (abs(tar-temp.val) < abs(tar-ans.val)) ans = temp;                                    store[temp].p1 = f[jj][j];                                    store[temp].p2 = f[kk][k];                                }                            }                        }                }            }        }        tot = 0;        if (store[ans].p1 != start || store[ans].p2 != start) {            tot++;            d[tot] = store[ans];            val[tot] = ans.val;            for (int i = 1; i <= tot; i++) {                if (store[d[i].p1].p1 != start) {                    tot++;                    d[tot] = store[d[i].p1];                    val[tot] = d[i].p1.val;                }                if (store[d[i].p2].p1 != start) {                    tot++;                    d[tot] = store[d[i].p2];                    val[tot] = d[i].p2.val;                }            }        }        printf("Target: %d\n", tar);        for (int i = tot; i >= 1; i--) {            //  +            if (d[i].p1.val + d[i].p2.val == val[i]) {                printf("%d + %d = %d\n", d[i].p1.val, d[i].p2.val, val[i]);                continue;            }            //  -            if (d[i].p1.val - d[i].p2.val == val[i]) {                printf("%d - %d = %d\n", d[i].p1.val, d[i].p2.val, val[i]);                continue;            }            if (d[i].p2.val - d[i].p1.val == val[i]) {                printf("%d - %d = %d\n", d[i].p2.val, d[i].p1.val, val[i]);                continue;            }            //  *            if (d[i].p1.val * d[i].p2.val == val[i]) {                printf("%d * %d = %d\n", d[i].p1.val, d[i].p2.val, val[i]);                continue;            }            //  /            if (d[i].p2.val != 0 && d[i].p1.val / d[i].p2.val == val[i]) {                printf("%d / %d = %d\n", d[i].p1.val, d[i].p2.val, val[i]);                continue;            }            if (d[i].p1.val != 0 && d[i].p2.val / d[i].p1.val == val[i]) {                printf("%d / %d = %d\n", d[i].p2.val, d[i].p1.val, val[i]);                continue;            }        }        printf("Best approx: %d\n", ans.val);    }}
0 0