E. Coins ——BNUOJ 规律题

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题目描述：

https://acm.bnu.edu.cn/v3/statments/52297.pdf

根据每个题目打表总结规律

a1 a2 a3 都不为0 为一种情况

其他在代码里展示：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)#define mod 2009#define INF 9999999999999#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}ll my_min(ll x,ll y){ return x>y?y:x;};int main(int argc,char *argv[]){ll a1,a2,a3;    while((scanf("%lld %lld %lld",&a1,&a2,&a3)!=EOF))    {    ll re = 0;        if(a1!=0&&a2==0&&a3==0)//一个不为0，其他两个为0     {    re = a1 ;    printf("%lld\n",re);    continue;}if(a1==0&&a2!=0&&a3==0)    {    re = a2 ;    printf("%lld\n",re);    continue;}    if(a1==0&&a2==0&&a3!=0)    {    re = a3;    printf("%lld\n",re);    continue;}        if(a1!=0&&a2!=0)    {    re = a1*1+a2*2+a3*3;}else if(a1 == 0 && a2 != 0){if(a2==1)//a2为1re = a3*2+1;else re = a3*3+2*a2-2; }else {if(a1==1)re = a3*2+1;else re = a3*3+a1;}printf("%lld\n",re);    }    return 0;}

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