div. 2
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n,m;
scanf("%d%d", &n, &m);
int last=0;
int ans=0;
while(n)
{
n--;
last++;
ans++;
if(last == m)
{
last = 0;
n++;
}
}
printf("%d\n",ans);
return 0;
}
#include <bits/stdc++.h>
using namespace std;
#define ll __int64
#define inf 1000000000
ll p[100];
int main()
{
int a, b, c, li=0;
scanf("%d%d%d", &a, &b, &c);
for(int i = 1; i <= 81; i++)
{
ll si = (ll)b * pow(i*1.0, a) + c;
int s=0;
ll sum = si;
while(sum)
{
s += sum%10;
sum/=10;
}
if(si > inf) continue;
if(s == i)
{
li++;
p[li] = si;
}
}
printf("%d\n", li);
for(int i = 1; i <= li; i++)
{
if(i!=li) printf("%I64d ", p[i]);
else printf("%I64d\n", p[i]);
}
return 0;
}n盆花，澆水可以澆m天，每次澆連續的w盆
每盆花有一個初始值a
澆一次水a加1
最終要使n盆花的最小值最大
求這個最大值
用折半查找的方法做
根據題目，我們可以知道a的範圍 1<=a<=10^9+10^5
折半查找a
#include <bits/stdc++.h>
using namespace std;
#define inf 100005
#define maxn 1e9+1e5
#define ll __int64
ll p[inf];
ll s[inf];
int n, m, w;
int solve(ll num)
{
memset(s, 0, sizeof(s));
ll last = 0;
ll wi = 0;
for(int i = 1; i <= n; i++)
{
last += s[i];
if(p[i] +last < num)
{
s[(i+w)>n?n+1:i+w] -= (num-p[i]-last);
wi += (num-p[i]-last);
last += (num-p[i]-last);
if(wi > (ll)m) return 0;
}
}
return 1;
}
int main()
{
scanf("%d%d%d", &n, &m, &w);
for(int i = 1; i <= n; i++)
{
scanf("%I64d", &p[i]);
}
ll l = 1, r = maxn;
ll ans = 0;
while(l <= r)
{
ll mid = (l+r) >> 1;
int s = solve(mid);
if(s == 1)
{
l = mid+1;
ans = max(ans , mid);
}
else
{
r = mid-1;
}
}
printf("%I64d\n", ans);
return 0;
}在l~r中，選出x個組成數列s[ ]（1<=x<=k），使得s[1]^s[2]^s[3]^……^s[x]最小
小知識點：(2*x) ^ (2*x+1) = 1
分別討論r-l+1 和 k
比較煩的就是s[]包含的元素只有3個的時候
3個元素的規律應當是這樣子的
1100……000
1011……111
0111……111
#include <bits/stdc++.h>
using namespace std;
#define ll __int64
int main()
{
ll l, r, k;
scanf("%I64d%I64d%I64d", &l, &r, &k);
if(k == 1 || r - l + 1 == 1)
{
printf("%I64d\n1\n%I64d\n", l, l);
return 0;
}
if(r - l + 1 == 2)
{
if(l&1)
{
ll ans = l^r;
if(l < ans) printf("%I64d\n1\n%I64d\n", l, l);
else printf("%I64d\n2\n%I64d %I64d\n", l^r, l, r);
}
else printf("1\n2\n%I64d %I64d\n", l, r);
return 0;
}
if(r - l + 1 == 3)
{
ll a, b;
if(l&1) a = l+1, b = r;
else a = l, b = l+1;
if(k == 2)
{
printf("1\n2\n%I64d %I64d\n", a, b);
return 0;
}
else
{
ll ans = l^(l+1);
ans = ans^r;
if(ans == 0)
{
printf("0\n3\n%I64d %I64d %I64d\n", l, l+1, r);
return 0;
}
else
{
printf("1\n2\n%I64d %I64d\n", a, b);
return 0;
}
}
}
if(r - l + 1 == 4)
{
if(k == 2)
{
ll a, b;
if(l&1) l++;
printf("1\n2\n%I64d %I64d\n", l, l+1);
return 0;
}
else
{
if(l&1 || k == 3)
{
ll a = l, b = l+1, c = l+2, d = l+3;
ll ans = a^b;
ans = ans^c;
if(ans == 0)
{
printf("0\n3\n%I64d %I64d %I64d\n", a, b, c);
return 0;
}
ans = a^b;
ans = ans^d;
if(ans == 0)
{
printf("0\n3\n%I64d %I64d %I64d\n", a, b, d);
return 0;
}
ans = a^c;
ans = ans^d;
if(ans == 0)
{
printf("0\n3\n%I64d %I64d %I64d\n", a, c, d);
return 0;
}
ans = b^c;
ans = ans^d;
if(ans == 0)
{
printf("0\n3\n%I64d %I64d %I64d\n", b, c, d);
return 0;
}
if(l&1) l++;
printf("1\n2\n%I64d %I64d\n", l, l+1);
}
else printf("0\n4\n%I64d %I64d %I64d %I64d\n", l, l+1, l+2, l+3);
}
return 0;
}
if(r - l + 1 > 4)
{
if(k == 2)
{
ll a, b;
if(l&1) a = l+1, b = l+2;
else a = l, b = l+1;
printf("1\n2\n%I64d %I64d", a, b);
return 0;
}
if(k == 3)
{
ll x=-1,y,z;
for(ll i=3; i<=r; i=i<<1)
{
if((i^(i-1))>=l)
{
x=i;
y=i - 1;
z=i^(i-1);
break;
}
}
if(x!=-1)
{
printf("0\n3\n%I64d %I64d %I64d\n",x,y,z);
}
else
{
if(l&1) l++;
printf("1\n2\n%I64d %I64d\n",l,l+1);
}
return 0;
}
if(k >= 4)
{
ll a, b, c, d;
if(l&1) l++;
printf("0\n4\n%I64d %I64d %I64d %I64d\n", l, l+1, l+2, l+3);
}
}
return 0;
}
