标签：它的 view limits sys inf pac name queue names
第一行一个整数n(1<=n<=1000000)，接下来一行n个数ai(1<=ai<=1000000)表示LYK的序列。
一行表示答案。
5 3 4 5 5 3
2
分析：区间[l,r]对x（l<=x<=r）的贡献次数为2*x-l-r;
所以维护前缀和，后缀和，然后对每个数算贡献即可；
注意取模2^32等价于unsigned long long的溢出；
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <unordered_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 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define Lson L, mid, ls[rt] #define Rson mid+1, R, rs[rt] #define sys system("pause") #define intxt freopen("in.txt","r",stdin) const int maxn=1e6+10; using namespace std; 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;} inline ll read() { ll x=0;int f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } int n,m,k,t,a[maxn]; ll suml[maxn],sumr[maxn],numl[maxn],numr[maxn],pos[maxn],pre[maxn],f[maxn],fl[maxn],fr[maxn]; unsigned ll ans; int main() { int i,j; scanf("%d",&n); rep(i,1,n)a[i]=read(); rep(i,1,n) { numl[i]=numl[pos[a[i]]]+1; suml[i]=suml[pos[a[i]]]+i; pos[a[i]]=i; } memset(pos,0,sizeof(pos)); for(i=n;i>=1;i--) { numr[i]=numr[pos[a[i]]]+1; sumr[i]=sumr[pos[a[i]]]+i; pos[a[i]]=i; } rep(i,1,n) { fl[i]=fl[i-1]; fl[i]+=i*numr[i]; fl[i]-=suml[i-1]; f[i]=f[i-1]+numr[i]-numl[i-1]; } for(i=n;i>=1;i--) { fr[i]=fr[i+1]; fr[i]+=i*numl[i]; fr[i]-=sumr[i+1]; ans=ans+a[i]*(2*i*f[i]-fl[i]-fr[i]); } printf("%u\n",ans); //system("Pause"); return 0; }
标签：它的 view limits sys inf pac name queue names
原文地址：http://www.cnblogs.com/dyzll/p/6128549.html
