爆long long处理方法

#include <iostream>
using namespace std;

typedef __int128_t i128;
typedef long long ll;

int main() {
	i128 ans = (i128)1<<100;
	ll mod = 1e18, x = ans%mod, y = ans/mod;
	if (y) printf("%lld%018lld\n",y,abs(x));
	else printf("%lld\n", x);
}

若只涉及相加操作, 并且每次相加的绝对值不超过1e18, 可以避免使用int128

#include <iostream>
using namespace std;

typedef __int128_t i128;
typedef long long ll;

int main() {
	ll mod = 1e18, x = 0, y = 0;
	for (int i=1; i<=100000; ++i) {
		ll t;
		cin>>t;
		x += t;
		if (x>=mod) x-=mod,++y;
		if (x<=-mod) x+=mod,--y;
	}
	if (x>0&&y<0) x-=mod,++y;
	if (x<0&&y>0) x+=mod,--y;
	if (y) printf("%lld%018lld\n",y,abs(x));
	else printf("%lld\n",x);
}