【回文串】 Jisuanke 商品推荐走马灯（中等）

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查询区间内子串是回文串的价值总和

先处理好 a[1]*1+a[2]*2+.....+   前缀和

以及价值的前缀和

然后减减加加就好了

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define clc(arr, val) memset(arr, val, sizeof(arr))
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define IN freopen ("in.txt" , "r" , stdin);
#define OUT freopen ("out.txt" , "w" , stdout);
typedef long long LL;
typedef unsigned long long ULL;
const int MAXN = 101000;
const int MAXM = 6000010;
const int INF = 0x3f3f3f3f;
const LL mod = (LL)1<<32;
const double eps= 1e-8;
const double pi=acos(-1.0);
#define lson l,m, rt<<1
#define rson m+1,r,rt<<1|1#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int val[MAXN];
int c[MAXN],p[MAXN];
LL fir[MAXN],fron[MAXN];
int solve(int len)
{
c[0]=999;
for(int i=0;i<len;i++)
c[i*2+1]=888,c[i*2+2]=val[i+1];
len=2*len+2;
c[len-1]=888;
c[len]=1234;
int mx=0,id=0;
for(int i=1;i<len;i++)
{
if(mx>i)
p[i]=min(p[2*id-i],p[id]+id-i);
else
p[i]=1;
while(c[i+p[i]]==c[i-p[i]])
p[i]++;
if(i+p[i]>mx)
mx=i+p[i],id=i;
}
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
fron[0]=fir[0]=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&val[i]);
fron[i]=fron[i-1]+(LL)i*val[i];
fir[i]=fir[i-1]+val[i];
}
solve(n);
int l,r;
while(m--)
{
scanf("%d%d",&l,&r);
LL ans=0;
for(int i=l*2;i<=r*2;i++)
{
int x=p[i]-1;
if(i&1)//偶串
{
int in=i/2;
x=x/2;
x=min(x,min(r-in,in-l+1));
ans+=(fron[in]-fron[in-x]-(LL)(in-x)*(fir[in]-fir[in-x]))*(LL)2;
}
else //奇串
{
int in=i/2;
x=x/2;
x=min(x,min(r-in,in-l));
ans+=(fron[in-1]-fron[in-x-1]-(LL)(in-x-1)*(fir[in-1]-fir[in-x-1]))*(LL)2+(fir[in]-fir[in-1])*(x+1);
}
}
printf("%lld\n",ans);
}
}
return 0;
}