题面
现场题面
codeforces878E
codeforces878E
A sequence of n integers is written on a blackboard. Soon Sasha will come to the blackboard and start the following actions: let x and y be two adjacent numbers ((x) before (y)), then he can remove them and write (x + 2y) instead of them. He will perform these operations until one number is left. Sasha likes big numbers and will get the biggest possible number.
Nikita wants to get to the blackboard before Sasha and erase some of the numbers. He has (q) options, in the option (i) he erases all numbers to the left of the (l_i)-th number and all numbers to the right of (r_i)-th number, i. e. all numbers between the (l_i)-th and the (r_i)-th, inclusive, remain on the blackboard. For each of the options he wants to know how big Sasha's final number is going to be. This number can be very big, so output it modulo (10^9 + 7).
双倍经验传送门：Luogu
题解
自己的唠叨
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现场得分：NULL/50（他没给我测）
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题目已补
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注意点：
- 会爆long long，所以对于一个块，要记录取模后的和以及大致的和（就是不取模，与INF取min）
- 一定要注意不要输出负数
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xjoi上面的数据是hapi，过了xjoi不一定能过codeforces。其中差别（就我的程序而言），在于为(0)的元素的处理。值为(0)的块不要并到前面的块上面。
现场题解
代码
#include<bits/stdc++.h>
#define LL long long
#define MAXN 1000000
#define MOD 1000000007
#define INF (1ll<<62)
using namespace std;
template<typename T>void Read(T &cn)
{
char c;int sig = 1;
while(!isdigit(c = getchar()))if(c == '-')sig = -1; cn = c-48;
while(isdigit(c = getchar()))cn = cn*10+c-48; cn*=sig;
}
template<typename T>void Write(T cn)
{
if(cn < 0) {putchar('-'); cn = 0-cn; }
int wei = 0; T cm = 0; int cx = cn%10; cn/=10;
while(cn)cm = cm*10+cn%10,cn/=10,wei++;
while(wei--)putchar(cm%10+48),cm/=10;
putchar(cx+48);
}
template<typename T>void Min(T &cn, T cm) {cn = cn > cm ? cm : cn; }
template<typename T>void Max(T &cn, T cm) {cn = cn < cm ? cm : cn; }
struct xun{
int l,r,ci;
inline friend bool operator <(xun a,xun b) {return a.r < b.r; }
void getit(int cn) {Read(l); Read(r); ci = cn; }
};
struct qwe{
LL zhi,dax,chang;
void mk(int cn) {zhi = dax = cn; }
};
xun b[MAXN+1];
int ans[MAXN+1];
int n,m;
qwe a[MAXN+1];
int fa[MAXN+1];
LL qz[MAXN+1],mi[MAXN+1],nim[MAXN+1];
LL he[MAXN+1];
int get_fa(int cn) {return cn == fa[cn] ? cn : fa[cn] = get_fa(fa[cn]); }
LL ksm(LL cn,LL cm) {LL ans = 1; while(cm) ans = ans*(1+(cn-1)*(cm&1))%MOD, cn = cn*cn%MOD, cm>>=1; return ans; }
void yuchu(int cn)
{
mi[0] = 1; mi[1] = 2; nim[0] = 1; nim[1] = ksm(2,MOD-2);
for(int i = 2;i<=cn;i++) mi[i] = mi[i-1]*mi[1]%MOD, nim[i] = nim[i-1]*nim[1]%MOD;
}
void weihu_he(int cn)
{
int lin = get_fa(cn);
he[lin] = (he[get_fa(lin-1)]+a[lin].zhi*(1+(lin!=1)))%MOD;
}
LL jia(LL cn,LL cm,LL cx)
{
if(cx >= 62) return INF;
if(((INF-cn)>>cx) <= cm) return INF;
LL guo = cn + cm*(1ll<<cx);
return min(guo,INF);
}
void weihu_a(int cn)
{
a[cn].chang = 1;
if(cn == 1) return;
while(cn != 1 && a[cn].dax >= 0)
{
int lin = get_fa(cn-1);
fa[cn] = lin;
a[lin].zhi = (a[lin].zhi + a[cn].zhi * mi[a[lin].chang])%MOD;
a[lin].dax = jia(a[lin].dax,a[cn].dax,a[lin].chang);
a[lin].chang = a[lin].chang + a[cn].chang;
cn = lin;
}
}
LL suan(int cn,int cm)
{
int lin = get_fa(cn);
int lin2 = lin + a[lin].chang-1;
int lin1 = get_fa(cm);
// printf("lin = %d lin1 = %d lin2 = %d
",lin,lin1,lin2);
return ((he[lin1] - he[lin] + (qz[lin2]-qz[cn-1]+MOD)%MOD*nim[cn-1]%MOD+MOD)%MOD+MOD)%MOD;
}
signed main()
{
// freopen("border2.in","r",stdin);
// freopen("T3.out","w",stdout);
Read(n); Read(m);
for(int i = 1;i<=n;i++) {int bx; Read(bx); a[i].mk(bx); fa[i] = i; }
for(int i = 1;i<=m;i++) b[i].getit(i);
yuchu(n); qz[0] = 0;
for(int i = 1;i<=n;i++) qz[i] = (qz[i-1] + a[i].zhi*mi[i-1])%MOD;
sort(b+1,b+m+1);
int xian = 0; he[0] = 0;
for(int i = 1;i<=n;i++)
{
weihu_a(i); weihu_he(i);
while(xian < m && b[xian+1].r == i) xian++, ans[b[xian].ci] = suan(b[xian].l,i);
}
for(int i = 1;i<=m;i++) Write(ans[i]), puts("");
return 0;
}