AtCoder Beginner Contest 186
A Brick
int n, m;
int main()
{
scanf("%d%d", &n, &m);
printf("%d\n", n / m);
return 0;
}
B Blocks on Grid
int n, m;
int a[N][N];
int main()
{
int sum = 0x3f3f3f3f;
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++ i)
for(int j = 1; j <= m; ++ j)
{
scanf("%d", &a[i][j]);
sum = min(sum, a[i][j]);
}
int res = 0;
for(int i = 1; i <= n; ++ i)
for(int j = 1; j <= m; ++ j)
{
res += a[i][j] - sum;
}
printf("%d\n", res);
return 0;
}
C Unlucky 7
bool check(int x)
{
int t = x;
while(x)
{
if(x % 10 == 7) return 1;
x /= 10;
}
while(t)
{
if(t % 8 == 7) return 1;
t /= 8;
}
return 0;
}
int main()
{
scanf("%d", &n);
int res = 0;
for(int i = 7; i <= n; ++ i)
if(check(i)) res ++;
printf("%d\n", n - res);
return 0;
}
D Sum of difference
int n;
int a[SZ], b[SZ];
int main()
{
LL sum = 0;
scanf("%d", &n);
for(int i = 1; i <= n; ++ i)
scanf("%d", &a[i]);
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; ++ i)
sum += (LL)a[i] * (2 * i - 1 - n);
printf("%lld\n", sum);
return 0;
}
E Throne
\(s + k * x = n * y\)
int n, s, k;
LL exgcd(LL a, LL b, LL &x, LL &y)
{
if(!b)
{
x = 1, y = 0;
return a;
}
LL d = exgcd(b, a % b, y, x);
y -= a / b * x;
return d;
}
int main()
{
int __;
scanf("%d", &__);
while(__ -- )
{
LL x, y;
scanf("%d%d%d", &n, &s, &k);
LL d = exgcd(k, n, x, y);
if(s % d) puts("-1");
else
{
x *= - s / d;
LL t = abs(n / d);
printf("%lld\n", (x % t + t) % t);
}
}
return 0;
}
F Rook on Grid
每一次移动可以往下或往右走任意步数,但遇到墙要停,初始位置在(1,1),问走两步以内能够覆盖的路径有多少个格子.
首先考虑第一步向下走,把第二步向右走可以到达的位置统计。
然后考虑第一步向右走,第二步向下走且第一种情况未到达的位置:
即到当前列第一个障碍这些行中,在前面列出现障碍的行数,统计完后把当前列的障碍更新上去,可以用一个树状数组维护.
int tr[N];
int n, m, k;
int row[N];
int column[N];
vector<int> v[N];
int lowbit(int x) { return x & -x; }
void add(int x, int v) { for(int i = x; i <= n; i += lowbit(i)) tr[i] += v; }
int query(int x) { int res = 0; for(int i = x; i; i -= lowbit(i)) res += tr[i]; return res; }
int main()
{
scanf("%d%d%d", &n, &m, &k);
for(int i = 1; i <= n; ++ i) row[i] = m + 1;
for(int i = 1; i <= m; ++ i) column[i] = n + 1;
for(int i = 1; i <= k; ++ i)
{
int x, y;
scanf("%d%d", &x, &y);
row[x] = min(row[x], y);
column[y] = min(column[y], x);
}
LL res = 0;
int flag = 0;
for(int i = 1; i <= n; ++ i)
{
if(flag == 1) { row[i] = 1; continue; }
if(row[i] == 1) { flag = 1; continue; }
res += row[i] - 1;
}
for(int i = 1; i <= n; ++ i) v[row[i]].push_back(i);
for(int i = 1; i <= m; ++ i)
{
if(column[i] == 1) break;
res += query(column[i] - 1);
for(auto j: v[i]) add(j, 1);
}
printf("%lld\n", res);
return 0;
}
2021.1.20
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