給你一個(gè)n個(gè)初始元素都為1的序列和m個(gè)詢(xún)問(wèn)q。
詢(xún)問(wèn)格式為：l r x(x為2or3)
最后求1~n所有數(shù)的GCD
GCD：把每個(gè)數(shù)分別分解質(zhì)因數(shù)，再把各數(shù)中的全部公有質(zhì)因數(shù)提取出來(lái)連乘，所得的積就是這幾個(gè)數(shù)的最大公約數(shù)。
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define mod 998244353#define pi acos(-1.0)
#define rep(i,x,n) for(int i=(x); i<(n); i++)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 1<<30;
const int maxn = 1e5+3;
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dir[2]={-1,1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
LL t,n,m;
LL Pow(LL a, LL b)
{LL res=1;while(b){if(b&1)res=(res%mod * a%mod)%mod;a = a%mod*a%mod;b>>=1;}return res%mod;
}
LL a[maxn],b[maxn];
int main()
{scanf("%lld",&t);while(t--){ms(a,0),ms(b,0);scanf("%lld%lld",&n,&m);while(m--){LL l,r,x;scanf("%lld%lld%lld",&l,&r,&x);if(x==2){a[l]++,a[r+1]--; //差分標(biāo)記因子含有2的數(shù)、區(qū)間加維護(hù)2or3的操作數(shù)}else{b[l]++,b[r+1]--;}}LL m1=a[1],m2=b[1];for(int i=2;i<=n;i++){a[i]+=a[i-1]; //前綴和維護(hù)序列本身，而序列記錄2的操作數(shù)即個(gè)數(shù)，得到具體每個(gè)數(shù)的操作數(shù)b[i]+=b[i-1];m1=min(m1,a[i]); //2的最小操作數(shù)m2=min(m2,b[i]);}LL ans = (Pow(2,m1)%mod*Pow(3,m2)%mod)%mod;printf("%lld\n",ans);}
}
/*
2
5 3
1 3 2
3 5 2
1 5 3
6 3
1 2 2
5 6 2
1 6 26 2
【題意】【類(lèi)型】【分析】【時(shí)間復(fù)雜度&&優(yōu)化】【trick】【數(shù)據(jù)】*/
轉(zhuǎn)載于:https://www.cnblogs.com/Roni-i/p/9527250.html
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