题目大意：给定一个由+1<script type="math/tex" id="MathJax-Element-517">+1</script>和?1<script type="math/tex" id="MathJax-Element-518">-1</script>构成的长度为n<script type="math/tex" id="MathJax-Element-519">n</script>的序列，提供两种操作：
1.将某一位取反，花销为x<script type="math/tex" id="MathJax-Element-520">x</script>
2.将最后一位移动到前一位。花销为y<script type="math/tex" id="MathJax-Element-521">y</script>
要求终于p+sumn=q<script type="math/tex" id="MathJax-Element-522">p+sum_n=q</script>。且p+sumi≥0(1≤i≤n)<script type="math/tex" id="MathJax-Element-523">p+sum_i\geq 0(1\leq i\leq n)</script>，求最小花销
枚举终于的序列以哪个点開始。那么从这个点往后的最小前缀和能够用单调队列预处理出来
然后贪心地把左边的?1<script type="math/tex" id="MathJax-Element-524">-1</script>改成+1<script type="math/tex" id="MathJax-Element-525">+1</script>。右边的+1<script type="math/tex" id="MathJax-Element-526">+1</script>改成?1<script type="math/tex" id="MathJax-Element-527">-1</script>直到满足要求就可以

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 2002002
using namespace std;
int n,p,q;
long long x,y,ans=0x7fffffffffffffffll;
char s[M];
int sum[M],_min[M];
void Assert(bool flag)
{
    if(flag)
        return ;
    printf("%d\n",1/0);
    exit(0);
}
int main()
{
    int i;
    cin>>n>>p>>q>>x>>y;
    scanf("%s",s+1);
    for(i=n<<1;i>n;i--)
        sum[i]=sum[i+1]+(s[i-n]==‘-‘?-1:1);
    for(i=n;i;i--)
        sum[i]=sum[i+1]+(s[i]==‘-‘?
-1:1);
    static int q[M],r,h;
    for(i=n<<1;i;i--)
    {
        while( r-h>=1 && sum[q[r]]<sum[i] )
            q[r--]=0;
        q[++r]=i;
        while( q[h+1]-i>n )
            q[++h]=0;
        if(i<=n)
            _min[i]=sum[i]-sum[q[h+1]];
    }
    Assert( ~ (::q)-p - sum[n+1] & 1 );
    int temp=(::q-p)-sum[n+1]>>1;
    for(i=1;i<=n;i++)
    {
        long long cost=(n-i+1)%n*y+abs(temp)*x;
        _min[i]+=p+max(temp,0)*2;
        if(_min[i]<0)
            cost+=(1-_min[i]>>1)*x*2;
        if(cost<ans)
            ans=cost;
    }
    cout<<ans<<endl;
    return 0;
}