2019ICPC银川区域赛 B. So Easy

链接

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题解

这种签到题得需要快速反应才行

因为每行的增量都是一定的，所以 $a_{i,j}-a_{i+1,j} = a_{i,j}-a_{i+1,k}$

也就是说对于某两行，你选出任意一列，得到的两个元素的差都是一样的

利用这个就能做此题

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
ll n, a[maxn][maxn], hi, hj;
int main()
{
    ll i, j;
    n = read();
    rep(i,1,n)rep(j,1,n)a[i][j]=read();
    rep(i,1,n)rep(j,1,n)if(a[i][j]==-1)hi=i, hj=j;
    if(hi<n)
    {
        rep(j,1,n)if(a[hi][j]!=-1)a[hi][hj] = a[hi+1][hj] + a[hi][j] - a[hi+1][j];
    }
    else
    {
        rep(j,1,n)if(a[hi][j]!=-1)a[hi][hj] = a[hi-1][hj] + a[hi][j] - a[hi-1][j];
    }
    printf("%lld",a[hi][hj]);
    return 0;
}