牛客练习赛8 储物点的距离

题目链接

 

当x<=l时 区间[l,r]到x的代价等于区间[l,r]到1的代价,减去重量[l,r]从1到x的代价

当x>=r时 区间[l,r]到x的代价等于区间[l,r]到n的代价,减去重量[l,r]从n到x的代价

其他 区间[l,r]到x的代价分成俩份[l,x-1]到x和[x+1,r]到x

 

用cost_l cost_r来维护到1和到n的花费,然后减去差的距离*[l,r]的数量就OK了。

 

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
//#pragma GCC optimize("Ofast")
#pragma comment(linker, "/STACK:102400000,102400000")
//#pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx) 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef long long LL;
typedef pair  ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=2e5+5;
const int maxx=1e7+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
templateinline T min(T a,T b,T c) { return min(min(a,b),c);}
templateinline T max(T a,T b,T c) { return max(max(a,b),c);}
templateinline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
templateinline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template 
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

int n,m;
LL sum[maxn];
LL l[maxn],a[maxn],b[maxn],r[maxn],cost_l[maxn],cost_r[maxn];
void init() {
    sum[0]=0;
    l[1]=0;
    r[n]=0;
    cost_r[n]=0;
    cost_l[1]=0;
}
void solve(){
    W(cin>>n>>m) {
        init();
        FOR(2,n,i) {
            S_1(a[i]);
            l[i]=(l[i-1]+a[i])%mod;
        }
        fOR(n-1,1,i){
            r[i]=(r[i+1]+a[i+1])%mod;
        }
        FOR(1,n,i) {
            S_1(b[i]);
            sum[i]=(sum[i-1]+b[i])%mod;
        }
        fOR(n-1,1,i) {//cost_r 鍒?
            cost_r[i]=(cost_r[i+1]+r[i]*b[i]%mod)%mod;
        }
        FOR(2,n,i){//cost_l 鍒皀
            cost_l[i]=(cost_l[i-1]+l[i]*b[i]%mod)%mod;
        }
        W(m--) {
            LL x,L,R;
            S_3(x,L,R);
            LL ans=0;
            if(x<=L) {
                ans=(cost_l[R]-cost_l[L-1]-(sum[R]-sum[L-1])*l[x]+mod)%mod;
            }
            else if(x>=R) { 
                ans=(cost_r[L]-cost_r[R+1]-(sum[R]-sum[L-1])*r[x]+mod)%mod;
            }
            else {
                ans=(cost_r[L]-cost_r[x]-(sum[x-1]-sum[L-1])*r[x]+mod)%mod;
                ans=(ans+(cost_l[R]-cost_l[x]-(sum[R]-sum[x])*l[x]+mod)%mod+mod)%mod;
            }
            print(ans);
        }
    }
}
int main() { 
    //freopen( "1.in" , "r" , stdin );
    //freopen( "1.out" , "w" , stdout );
    int t=1;
    //init();
    //s_1(t);
    for(int cas=1;cas<=t;cas++) {
        //printf("Case #%d: ",cas);
        solve();
    }
}

 

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