寒假训练第四天-Educational Codeforces Round 102 (Rated for Div. 2)
前言：打了场div2，心态爆炸 ，ed场教做人，c读半天没读懂，d被memset卡死了， 菜的抠脚
题目链接-https://codeforces.com/contest/1473
A-Replacing Elements
题意：给你一个序列和一个限制d，问是否存在j， k, 使得任意 a[i] = a[j] + a[k] (i≠j; i≠k; j≠k) ，若a[i] <=d,则无需替换 。
题解：找出序列中最小的两个数，判断其和是否不大于d，当然若整个序列都没有大于d的，那就无需替换，直接输出yes即可。
int a[110], n, d;
int32_t main()
{
ICO;
int t;
cin >> t;
while(t--)
{
cin >> n >> d;
bool ok = 0;
for(int i = 1; i <= n; i++)
{
cin >> a[i];
if(a[i] > d) ok = 1;
}
if(!ok)
{
cout << "YES" << endl;
continue;
}
sort(a + 1, a + n + 1);
if(a[1] + a[2] <= d) cout << "YES" << endl;
else cout << "NO" << endl;
}
return 0;
}
B-String LCM
题意：给你两个字符串，求这两个字符串的最小公倍数。
题解：求出两个字符串长度的最小公倍数，然后将两个字符串增加，看其是否相等。
string s, t;
int32_t main()
{
ICO;
int q;
cin >> q;
while(q--)
{
cin >> s >> t;
int n = s.size(), m = t.size();
int p = n * m / __gcd(n, m);
string ss = s, tt = t;
for(int i = 1; i < p / n; i++)
ss += s;
for(int i = 1; i < p / m; i++)
tt += t;
if(ss == tt) cout << ss << endl;
else cout << -1 << endl;
}
return 0;
}
D- Program
题解：用前缀和将舍弃的一段算出来，然后用st表维护两端区间的最值， 将右边的区间的最值减去舍弃的区间的值，则左边的最值和右边的最值肯定有交集，找出两区间的最小值和最大值即可，判断一下0是否在区间里。
int n, m;
string s;
int d[maxn][25], a[maxn], Log[maxn], v[maxn][25];
void init()
{
for (int i = 1; i <= n; i++)
v[i][0] = d[i][0] = a[i];
for (int j = 1; (1 << j) <= n; j++)
{
for (int i = 1; i + (1 << j - 1) <= n; i++)
{
d[i][j] = max(d[i][j - 1], d[i + (1 << j - 1)][j - 1]);
v[i][j] = min(v[i][j - 1], v[i + (1 << j - 1)][j - 1]);
}
}
for (int i = 1; i <= n; i++)
Log[i] = log2(i);
}
int query_min(int l, int r)
{
int k = Log[r - l + 1];
return min(v[l][k], v[r - (1 << k) + 1][k]);
}
int query_max(int l, int r)
{
int k = Log[r - l + 1];
return max(d[l][k], d[r - (1 << k) + 1][k]);
}
int32_t main()
{
ICO;
int q, b[5], res, l , r;;
cin >> q;
while(q--)
{
cin >> n >> m;
cin >> s;
s = "0" + s;
for(int i = 1; i <= n; i++)
{
if(s[i] == '+') a[i] = a[i - 1] + 1;
else a[i] = a[i - 1] - 1;
}
init();
while(m--)
{
cin >> l >> r;
int p = a[r] - a[l - 1];
if(l == 1) b[1] = b[2] = 0;
else b[1] = query_min(1, l - 1), b[2] = query_max(1, l - 1);
if(r == n) b[3] = b[4] = p;
else b[3] = query_min(r + 1, n), b[4]= query_max(r + 1, n);
b[3] -= p, b[4] -= p;
sort(b + 1, b + 5);
res = b[4] - b[1] + 1;
if(b[1] > 0 || b[4] < 0) res++;
cout << res << endl;
}
memset(a, 0, 4 * n + 4 * 5);
for (int j = 1; (1 << j) <= n; j++)
for (int i = 1; i + (1 << j - 1) <= n; i++)
d[i][j] = v[i][j] = 0;
}
return 0;
}
总结：自闭自闭