# # AtCoder Beginner Contest 185 题解

## # Problem A - ABC Preparation (opens new window)

print(min(map(int, input().split())))

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## # Problem B - Smartphone Addiction (opens new window)

n, m, t = map(int, input().split())
last = 0
now = n
for _ in range(m):
ai, bi = map(int, input().split())
now -= ai - last
if now <= 0:
print('No')
exit(0)
now += bi - ai
if now > n:
now = n
last = bi
now -= t - last
print('Yes' if now > 0 else 'No')

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## # Problem C - Duodecim Ferra (opens new window)

from math import comb

l = int(input())
print(comb(l - 1, 11))

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## # Problem D - Stamp (opens new window)

n, m = map(int, input().split())
a = [] if m == 0 else list(map(int, input().split()))
a.sort()
if m == 0:
print(1)
else:
seg = []
if a[0] != 1:
seg.append(a[0] - 1)
if a[-1] != n:
seg.append(n - a[-1])
for i in range(m - 1):
if a[i + 1] - a[i] > 1:
seg.append(a[i + 1] - a[i] - 1)
if len(seg) == 0:
print(0)
else:
shortest = min(seg)
print(sum((si - 1) // shortest + 1 for si in seg))

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## # Problem E - Sequence Matching (opens new window)

n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
dp = [[0 for _ in range(m + 1)] for _ in range(n + 1)]
for i in range(n + 1):
for j in range(m + 1):
if i == 0 and j == 0:
continue
dp[i][j] = int(1e6)
if i > 0:
dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1)
if j > 0:
dp[i][j] = min(dp[i][j], dp[i][j - 1] + 1)
if i > 0 and j > 0:
extra = -1 if a[i - 1] == b[j - 1] else 0
dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + 1 + extra)
print(dp[n][m])

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## # Problem F - Range Xor Query (opens new window)

#include <atcoder/segtree>
#include <iostream>
#include <vector>

using namespace std;

int op(int a, int b) { return a ^ b; }

int e() { return 0; }

int main() {
int n, q;
cin >> n >> q;
vector<int> v(n);
for (int i = 0; i < n; ++i)
cin >> v[i];
atcoder::segtree<int, op, e> seg(v);
while (q--) {
int t, x, y;
cin >> t >> x >> y;
if (t == 1) {
seg.set(x - 1, v[x - 1] ^ y);
v[x - 1] ^= y;
} else {
cout << seg.prod(x - 1, y) << endl;
}
}
}
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