source scanf borde array size ltr d+ miss !=

大神題解:

http://blog.csdn.net/u014800748/article/details/47680899

The sum of gcd

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 526 Accepted Submission(s): 226


Problem Description You have an array ,the length of is
Let

Input There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
First line has one integers
Second line has integers
Third line has one integers ,the number of questions
Next there are Q lines,each line has two integers ,




Output For each question,you need to print
Sample Input

2
5
1 2 3 4 5
3
1 3
2 3
1 4
4
4 2 6 9
3
1 3
2 4
2 3

Sample Output

9
6
16
18
23
10

Author SXYZ
Source 2015 Multi-University Training Contest 8

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

const int maxn=10100;
typedef long long int LL;

struct G
{
    G(){}
    G(int _id,LL _g):id(_id),g(_g){}

    int id;
	LL g;

	void toString()
	{
		printf("id: %d g: %lld\n",id,g);
	}
};

int n,a[maxn],Q;
vector<G> VL[maxn],VR[maxn];
struct Que
{
    int L,R,id;
    bool operator<(const Que& que) const
    {
        if(L!=que.L) return L<que.L;
        return R<que.R;
    }
}que[maxn];

void PreInit()
{
    /// get Left Point
    /// 以i為右端點,預處理出左邊的段
    for(int i=1;i<=n;i++)
    {
        VL[i].clear();
        if(i==1)
        {
            VL[i].push_back(G(i,a[i]));
        }
        else
        {
			LL curg=a[i];int L=i;
			for(auto &it : VL[i-1])
			{
				int g=__gcd(it.g,curg);
				if(g!=curg) VL[i].push_back(G(L,curg));
				curg=g; L=it.id;
			}
			VL[i].push_back(G(L,curg));
        }
    }
    /// get Right Point
    /// 以i為左端點,預處理出右邊的段
	for(int i=n;i>=1;i--)
	{
		VR[i].clear();
		if(i==n)
		{
			VR[i].push_back(G(i,a[i]));
		}
		else
		{
			LL curg=a[i];int R=i;
			for(auto &it : VR[i+1])
			{
				int g=__gcd(curg,it.g);
				if(g!=curg) VR[i].push_back(G(R,curg));
				curg=g; R=it.id;
			}
			VR[i].push_back(G(R,curg));
		}
	}
}

/// 計算L,R之間的值
LL calu(int type,int L,int R)
{
	LL ret=0;
	if(type==0)
	{
		int tr=R;
		for(auto &it : VL[R])
		{
			if(it.id>=L)
			{
				ret+=(tr-it.id+1)*it.g;
				tr=it.id-1;
			}
			else
			{
				ret+=(tr-L+1)*it.g;
				break;
			}
		}
	}
	else if(type==1)
	{
		int tr=L;
		for(auto &it : VR[L])
		{
			if(it.id<=R)
			{
				ret+=(it.id-tr+1)*it.g;
				tr=it.id+1;
			}
			else
			{
				ret+=(R-tr+1)*it.g;
				break;
			}
		}
	}
	return ret;
}

LL ans[maxn];

int main()
{
	int T_T;
	scanf("%d",&T_T);
    while(T_T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        PreInit();

		scanf("%d",&Q);
        for(int i=0,l,r;i<Q;i++)
        {
            scanf("%d%d",&l,&r);
            que[i].L=l; que[i].R=r; que[i].id=i;
        }
        sort(que,que+Q);

		int L=1,R=0; LL ret=0;
		for(int i=0;i<Q;i++)
		{
			while(R<que[i].R)
			{
				R++;
				ret+=calu(0,L,R);
			}
			while(R>que[i].R)
			{
				ret-=calu(0,L,R);
				R--;
			}
			while(L<que[i].L)
			{
				ret-=calu(1,L,R);
				L++;
			}
			while(L>que[i].L)
			{
				L--;
				ret+=calu(1,L,R);
			}
			ans[que[i].id]=ret;
		}

		for(int i=0;i<Q;i++)
			cout<<ans[i]<<endl;
    }
    return 0;
}

HDOJ 5381 The sum of gcd 莫隊算法