7-10 5个砝码 (25 分)

本文介绍了一种使用有限数量的砝码通过组合称出特定重量的算法。以5个砝码为例,重量分别为1,3,9,27,81,能够组合称出1到121之间的任意整数重量。文章提供了AC代码实现,详细解释了如何通过枚举每个砝码的三种状态(不选、加、减)来找出符合目标重量的砝码组合。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

题目描述:

用天平称重时,我们希望用尽可能少的砝码组合称出尽可能多的重量。如果只有5个砝码,重量分别是1,3,9,27,81。则它们可以组合称出1到121之间任意整数重量(砝码允许放在左右两个盘中)。
输入格式:
本题目要求编程实现:对用户给定的重量,给出砝码组合方案。
输出格式:
emmmm看样例

输入样例:
在这里给出一组输入。例如:
5
19
输出样例:
在这里给出相应的输出。例如:
9-3-1
27-9+1

分析:

对于每一个数只有三种可能性。即,这个数存在不选,+,-这三种情况。
那么我们对于每一种情况都枚举一遍就可以了。
参考博客:https://blog.youkuaiyun.com/Sharing_Li/article/details/8760730
AC代码:

#include"stdio.h"
#include"string.h"
#include"algorithm"
using namespace std;
int main()
{
    int a[3]={0,1,-1};
    int b[5]={0,3,-3};
    int c[3]={0,9,-9};
    int d[3]={0,27,-27};
    int e[3]={0,81,-81};
    int digit[5]={0};
    int n;
    scanf("%d",&n);
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
            for(int k=0;k<3;k++)
               for(int x=0;x<3;x++)
                   for(int y=0;y<3;y++)
                      if(a[i]+b[j]+c[k]+d[x]+e[y]==n)
                      {
                            digit[0]=a[i];digit[1]=b[j];
                            digit[2]=c[k];digit[3]=d[x];
                            digit[4]=e[y];
                      }
    }

    int mark=1;
    for(int i=4;i>=0;i--)//从后往前输出
    {
        if(digit[i]==0)
            continue;
        if(mark==1)
            {
                printf("%d",digit[i]);mark=0;continue;
            }
        if(digit[i]>0)
        {
            printf("+%d",digit[i]);
        }
        else
            printf("%d",digit[i]);
    }
    printf("\n");
}
VBA 解析下列字符串 {"data":{"rows":[{"_id":"6854bd993061d70c6c87aa4d","68400066ea04d23da81f1953":"","rowid":"737888fe-28d5-4652-8eab-d993f22e08e7","ctime":"2025-06-20 09:47:05","caid":{"accountId":"00b093d0-a7bb-4b31-af89-eaaaf45defed","fullname":"肖飞龙","avatar":"https://quickapp.wuxiapptec.com/file/mdpic/UserAvatar/default12.png?watermark/2/text/6aOe6b6Z/font/5oCd5rqQ6buR5L2T/fontsize/700/fill/d2hpdGU=/dissolve/100/gravity/Center/dx/0/dy/0/fontstyle/Ym9sZA==%7CimageView2/1/w/48/h/48/q/90","isPortal":false,"status":1},"uaid":{"accountId":"user-workflow","fullname":"工作流","avatar":"https://quickapp.wuxiapptec.com/file/mdpic/UserAvatar/workflow.png?imageView2/1/w/48/h/48/q/90","isPortal":false,"status":1},"ownerid":{"accountId":"00b093d0-a7bb-4b31-af89-eaaaf45defed","fullname":"肖飞龙","avatar":"https://quickapp.wuxiapptec.com/file/mdpic/UserAvatar/default12.png?watermark/2/text/6aOe6b6Z/font/5oCd5rqQ6buR5L2T/fontsize/700/fill/d2hpdGU=/dissolve/100/gravity/Center/dx/0/dy/0/fontstyle/Ym9sZA==%7CimageView2/1/w/48/h/48/q/90","i sPortal":false,"status":1},"utime":"2025-06-20 10:32:24","683fff9fea04d23da81f1948":"","6840021fea04d23da81f19f3":"正常","684bb415ea04d23da823b097":"Non-ISO 9001","6840021fea04d23da81f19f4":["Agilent"],"683ff8bf5c36bb7f2c7478a7":"测试","6840021fea04d23da81f19f5":"[{\"accountId\":\"00b093d0-a7bb-4b31-af89-eaaaf45defed\",\"fullname\":\"肖飞龙\",\"avatar\":\"https://quickapp.wuxiapptec.com/file/mdpic/UserAvatar/default12.png?watermark/2/text/6aOe6b6Z/font/5oCd5rqQ6buR5L2T/fontsize/700/fill/d2hpdGU=/dissolve/100/gravity/Center/dx/0/dy/0/fontstyle/Ym9sZA==%7CimageView2/1/w/48/h/48/q/90\",\"status\":1}]","684003a1ea04d23da81f1a06":null,"684b81a1ea04d23da823a915":"[]","684b8370ea04d23da823a937":"[]","684f99d55c44ea7978e3e365":"[]","683ffc8fea04d23da81f1911":"[{\"type\":0,\"sid\":\"ffcb2f30-9aff-4d0b-914f-9c1202cf128e\",\"sidext\":\"\",\"accountId\":\"\",\"fullname\":\"\",\"avatar\":\"\",\"name\":\"液相色谱质联用仪\",\"ext1\":\"\",\"ext2\":\"\",\"link\":\"/worksheet/684be40b921b18b25df12b70/row/ffcb2f30-9aff-4d0b-91 4f-9c1202cf128e\",\"projectId\":\"\",\"sourcevalue\":\"{\\\"_id\\\":\\\"684be51ade06b48f5a0421ff\\\",\\\"wsid\\\":\\\"684be40b921b18b25df12b70\\\",\\\"rowid\\\":\\\"ffcb2f30-9aff-4d0b-914f-9c1202cf128e\\\",\\\"status\\\":1,\\\"683febc45c36bb7f2c7478a0\\\":\\\"液相色谱质联用仪\\\",\\\"unreads\\\":false,\\\"autoid\\\":19,\\\"discussunreads\\\":false,\\\"allowedit\\\":false,\\\"allowdelete\\\":false,\\\"controlpermissions\\\":\\\"\\\"}\"}]","683ffd3cea04d23da81f192b":"液相色谱质联用仪","683ffd3cea04d23da81f192c":"LCMS","683ffc8fea04d23da81f1913":"C","684f69245c44ea7978e3df4a":["记录","维护","维修","校验"],"68400066ea04d23da81f1952":"","autoid":0,"685138403061d70c6c8781f5":"https://quickapp.wuxiapptec.com/app/f934aaa4-2f56-41c8-9bd2-ca728bb1086d/684be40b921b18b25df12b6f/684be40b921b18b25df12b73/row/737888fe-28d5-4652-8eab-d993f22e08e7","683ff8bf5c36bb7f2c7478a5":"CAS-CD-LCMS-AZ","683fff3eea04d23da81f193c":"","683ffc8fea04d23da81f1914":"","683ff8bf5c36bb7f2c7478aa":"","6853d5645c44ea7978e479fa":"","6853d47da716e0 89c0026387":"","68520f2f5c44ea7978e46803":"","6855075c5c44ea7978e48224":"[]","6853d5645c44ea7978e479fb":"","6853d5645c44ea7978e479f9":"","6852197b3061d70c6c87886a":"","684fb9d43061d70c6c856a66":"[]","6840021fea04d23da81f19f6":""},{"_id":"684bea4c6b544ef97b534f82","rowid":"e46f1378-b08f-4236-82f6-dc8d908d9565","ctime":"2025-06-13 17:07:24","caid":{"accountId":"00b093d0-a7bb-4b31-af89-eaaaf45defed","fullname":"肖飞龙","avatar":"https://quickapp.wuxiapptec.com/file/mdpic/UserAvatar/default12.png?watermark/2/text/6aOe6b6Z/font/5oCd5rqQ6buR5L2T/fontsize/700/fill/d2hpdGU=/dissolve/100/gravity/Center/dx/0/dy/0/fontstyle/Ym9sZA==%7CimageView2/1/w/48/h/48/q/90","isPortal":false,"status":1},"uaid":{"accountId":"user-workflow","fullname":"工作流","avatar":"https://quickapp.wuxiapptec.com/file/mdpic/UserAvatar/workflow.png?imageView2/1/w/48/h/48/q/90","isPortal":false,"status":1},"ownerid":{"accountId":"00b093d0-a7bb-4b31-af89-eaaaf45defed","fullname":"肖飞龙","avatar":"https://quickapp.wuxiapptec.com/file/mdpic/UserAva tar/default12.png?watermark/2/text/6aOe6b6Z/font/5oCd5rqQ6buR5L2T/fontsize/700/fill/d2hpdGU=/dissolve/100/gravity/Center/dx/0/dy/0/fontstyle/Ym9sZA==%7CimageView2/1/w/48/h/48/q/90","isPortal":false,"status":1},"utime":"2025-06-21 14:42:45","683ff8bf5c36bb7f2c7478a5":"CAS-CD-WEI-I","6840021fea04d23da81f19f3":"正常","684bb415ea04d23da823b097":"ISO 9001","683ff8bf5c36bb7f2c7478a7":"2mg (E2)","683fff3eea04d23da81f193c":"2025-04-27","683fff9fea04d23da81f1948":"2026-04-26","6840021fea04d23da81f19f6":"[]","684003a1ea04d23da81f1a06":"ADQC","6840021fea04d23da81f19f4":["其他:水玲砝码"],"684b8370ea04d23da823a937":"[]","683ffc8fea04d23da81f1911":"[{\"type\":0,\"sid\":\"174d5754-5277-49b2-8bb7-2dc225f8730d\",\"sidext\":\"\",\"accountId\":\"\",\"fullname\":\"\",\"avatar\":\"\",\"name\":\"砝码\",\"ext1\":\"\",\"ext2\":\"\",\"link\":\"/worksheet/684be40b921b18b25df12b70/row/174d5754-5277-49b2-8bb7-2dc225f8730d\",\"projectId\":\"\",\"sourcevalue\":\"{\\\"_id\\\":\\\"684be961a716e089c0017ffe\\\",\\\"wsid\\\":\\\"684be40b921b1 8b25df12b70\\\",\\\"rowid\\\":\\\"174d5754-5277-49b2-8bb7-2dc225f8730d\\\",\\\"status\\\":1,\\\"683febc45c36bb7f2c7478a0\\\":\\\"砝码\\\",\\\"unreads\\\":false,\\\"autoid\\\":0,\\\"discussunreads\\\":false,\\\"allowedit\\\":false,\\\"allowdelete\\\":false,\\\"controlpermissions\\\":\\\"\\\"}\"}]","683ffd3cea04d23da81f192b":"砝码","683ffd3cea04d23da81f192c":"WEI","683ffc8fea04d23da81f1913":"AE","684b81a1ea04d23da823a915":"[]","68400066ea04d23da81f1953":"","autoid":203,"684f69245c44ea7978e3df4a":["校验"],"685138403061d70c6c8781f5":"https://quickapp.wuxiapptec.com/app/f934aaa4-2f56-41c8-9bd2-ca728bb1086d/684be40b921b18b25df12b6f/684be40b921b18b25df12b73/row/e46f1378-b08f-4236-82f6-dc8d908d9565","6855075c5c44ea7978e48224":"[\"9219053d-1d51-40cd-b664-dea125ed89a5\"]","6840021fea04d23da81f19f5":"","68400066ea04d23da81f1952":"","683ffc8fea04d23da81f1914":"","683ff8bf5c36bb7f2c7478aa":"","6853d5645c44ea7978e479fa":"","6853d47da716e089c0026387":"","684f99d55c44ea7978e3e365":"[]","68520f2f5c44ea7978e46803":"","6853d56 45c44ea7978e479fb":"","6853d5645c44ea7978e479f9":"","6852197b3061d70c6c87886a":"","684fb9d43061d70c6c856a66":"[]"}],"total":204},"success":true,"error_code":1}
最新发布
06-23
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值