思路

mod everything

易错点1：单个值也要取模！！因为所有输入值最大都是1e18,所以每个值都要取模

易错点2：负数要加mod(a * b * c取模后可能会小于d * e * f)

易错点3：两两相乘时也要取模

#include<bits/stdc++.h>

using namespace std;
const int mod=998244353;
int main(){
	freopen("abc.in","r",stdin);
	freopen("abc.out","w",stdout);
	long long a,b,c,d,e,f;
	cin>>a>>b>>c>>d>>e>>f;
	long long cnt1=(((a%mod)*(b%mod)%mod)*(c%mod))%mod;
	long long cnt2=(((d%mod)*(e%mod)%mod)*(f%mod))%mod;
	long long res=((cnt1-cnt2)+mod)%mod;
	cout<<res;
	return 0;
} 

#include <cstdio>
#define MOD 998244353

// hack: a * b * c % MOD < d * e * f % MOD

long long a, b, c, d, e, f, abc_sum, def_sum, ans;
FILE * fin, * fout;

long long mul_mod(long long a, long long b, long long c)
{
    return (a % MOD) * (b % MOD) % MOD * (c % MOD) % MOD;
}

int main()
{
    fin = fopen("abc.in", "rb");
    fout = fopen("abc.out", "wb");

    fscanf(fin, "%lld %lld %lld %lld %lld %lld", &a, &b, &c, &d, &e, &f);

    abc_sum = mul_mod(a, b, c);
    def_sum = mul_mod(d, e, f);

    ans = (abc_sum - def_sum + MOD) % MOD;

    fprintf(fout, "%lld", ans);

    return 0;
}