C++14(g++5.4) 解法, 执行用时: 284ms, 内存消耗: 79212K, 提交时间: 2019-12-14 14:02:55

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 998244353;
const int N = 2010;
int A[N][N], B[N*2][N*2];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int n; cin >> n;
    for(int i = 0; i < n; i++) {
        int a, b; cin >> a >> b;
        A[a][b]++;
    }
    for(int i = 2000; i >= 0; i--) {
        for(int j = 2000; j >= 0; j--) {
            A[i][j] = (A[i][j] + A[i+1][j]) % mod;
            A[i][j] = (A[i][j] + A[i][j+1]) % mod;
            A[i][j] = (A[i][j] - A[i+1][j+1]) % mod;
        }
    }
    for(int i = 0; i <= 2000; i++) {
        for(int j = 0; j <= 2000; j++) {
            B[2000-i][2000-j] = A[i][j];
        }
    }
    int ans = 0;
    for(int i = 0; i <= 4000; i++) {
        for(int j = 0; j <= 4000; j++) {
            if(!i && !j) continue;
            if(i) B[i][j] = (B[i][j] + B[i-1][j]) % mod;
            if(j) B[i][j] = (B[i][j] + B[i][j-1]) % mod;
            if(i >= 2000 && j >= 2000) {
                ans = (ans + (ll) B[i][j] * A[i-2000][j-2000]) % mod;
            }
        }
    }
    ans = (ans - (ll) A[0][0] * A[0][0]) % mod;
    if(ans < 0) ans += mod;
    cout << ans << '\n';
    return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 566ms, 内存消耗: 168556K, 提交时间: 2019-12-14 12:07:25

#include<bits/stdc++.h>
#define ll long long
using namespace std;
struct aaa{
	ll x,y;
}a[200010];
ll n,i,j,x,y,ans,f[4010][4010],p=998244353,b[4010][4010];
int main(){
	scanf("%lld",&n);
	for(i=1;i<=n;i++){
		scanf("%lld%lld",&a[i].x,&a[i].y);
		b[a[i].x][a[i].y]++;
	}
	for(i=2000;i>=0;i--){
		for(j=2000;j>=0;j--){
			b[i][j]+=b[i+1][j]+b[i][j+1]-b[i+1][j+1];
			//if(b[i][j])printf("%lld %lld %lld\n",i,j,b[i][j]);
		}
	}
	for(i=-2000;i<=2000;i++){
		for(j=-2000;j<=2000;j++){
			x=i+2000;y=j+2000;
			if(x>0)f[x][y]+=f[x-1][y];
			if(y>0)f[x][y]+=f[x][y-1];
			if(i<=0&&j<=0)f[x][y]+=b[-i][-j];f[x][y]%=p;
			if(i>=0&&j>=0)ans=(ans+f[x][y]*b[i][j])%p;
			//if(f[x][y]&&b[i][j])printf("%lld %lld %lld %lld\n",i,j,b[i][j],f[x][y]);
		}
	}
	ans=(ans-n*n%p+p)%p;
	printf("%lld",ans);
}