做法:dp

思路:

因为音量范围在0~maxx之间并且maxx<=1000,所以我们设dp[i][j]为第i次音量变化后此时音量为j的方案是否存在

在符合条件的音量内进行转移

代码

// Problem: [HAOI2012]音量调节
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/problem/19990
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp(aa,bb) make_pair(aa,bb)
#define _for(i,b) for(int i=(0);i<(b);i++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,b,a) for(int i=(b);i>=(a);i--)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define X first
#define Y second
#define lowbit(a) (a&(-a))
#define debug(a) cout<<#a<<":"<<a<<"\n"
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef long double ld;
const int N=100010;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const double eps=1e-6;
const double PI=acos(-1.0);

int c[55];
bool dp[55][1005];

void solve(){
    mst(dp,0);
    int n,begin,maxx;cin>>n>>begin>>maxx;
    rep(i,1,n) cin>>c[i];
    dp[0][begin]=1;
    rep(i,1,n){
        rep(j,0,maxx){
            if(j-c[i]>=0&&dp[i-1][j-c[i]]) dp[i][j]=1;
            if(j+c[i]<=maxx&&dp[i-1][j+c[i]]) dp[i][j]=1;
        }
    }
    per(i,maxx,0){
        if(dp[n][i]) return cout<<i<<"\n",void(0);
    }
    return cout<<"-1\n",void(0);
}


int main(){
    ios::sync_with_stdio(0);cin.tie(0);
//    int t;cin>>t;while(t--)
    solve();
    return 0;
}