A,B,C都是水题。。。

D题,直接爆搜。我换了好多姿势,其实最简单的方法,就能过。

#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
using namespace std;
int que[];
int p[];
int o[];
int temp[];
int st[];
int ans,minz;
int n,a,b;
void dfs(int x,int sum)
{
int i;
if(sum >= minz) return;
if(x == n)
{
for(i = ;i <= n;i ++)
temp[i] = p[i];
for(i = ;i <= n-;i ++)
{
temp[i] -= o[i]*a;
temp[i-] -= o[i]*b;
temp[i+] -= o[i]*b;
}
for(i = ;i <= n-;i ++)
{
if(temp[i] >= ) break;
}
if(i == n)
{
if(sum < minz)
{
minz = sum;
for(i = ;i <= n-;i ++)
st[i] = o[i];
}
}
return ;
}
for(i = ;i <= ;i ++)
{
o[x] = i;
dfs(x+,sum+i);
}
}
int main()
{
int i,j;
scanf("%d%d%d",&n,&a,&b);
for(i = ; i <= n; i ++)
{
scanf("%d",&p[i]);
}
while(p[] >= )
{
que[ans++] = ;
p[] -= b;
p[] -= a;
p[] -= b;
}
while(p[n] >= )
{
que[ans++] = n-;
p[n-] -= b;
p[n] -= b;
p[n-] -= a;
}
for(i = ;i <= n;i ++)
{
if(p[i] >= ) break;
}
if(i == n+)
{
printf("%d\n",ans);
for(i = ;i < ans;i ++)
printf("%d ",que[i]);
printf("\n");
return ;
}
minz = ;
dfs(,);
printf("%d\n",minz+ans);
for(i = ;i < ans;i ++)
printf("%d ",que[i]);
for(i = ;i <= n-;i ++)
{
for(j = ;j <= st[i];j ++)
printf("%d ",i);
}
printf("\n");
return ;
}

E题,二分+rmq之类的,能求区间最值的算法。

题意是比较扯,按章节排列的n本书,要展示,最长能展示多少本,这一段的高度差要小于等于k,再输出总共多少中情况,把起点,终点输出。

rmq好久没写过了啊,记不得 什么东西了。

#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
using namespace std;
int n,k;
int p[];
int dpmin[][];
int dpmax[][];
int bin[];
void CL(int n)
{
int i,j;
for(i = ;i <= n;i ++)
{
dpmin[][i] = p[i];
dpmax[][i] = p[i];
}
for(i = ;bin[i] <= n;i ++)
{
for(j = ;j + bin[i-] <= n;j ++)
{
dpmin[i][j] = min(dpmin[i-][j],dpmin[i-][j+bin[i-]]);
dpmax[i][j] = max(dpmax[i-][j],dpmax[i-][j+bin[i-]]);
}
}
}
int rmqmax(int s,int t)
{
int k = log((t-s+)*1.0)/log(2.0);
return max(dpmax[k][s],dpmax[k][t-bin[k]+]);
}
int rmqmin(int s,int t)
{
int k = log((t-s+)*1.0)/log(2.0);
return min(dpmin[k][s],dpmin[k][t-bin[k]+]);
}
int query(int s,int t)
{
return rmqmax(s,t) - rmqmin(s,t);
}
int judge(int mid)
{
int i,ans = ;
for(i = ;i <= n-mid+;i ++)
{
if(query(i,i+mid-) <= k)
ans ++;
}
return ans;
}
int main()
{
int i,j,str,end,mid;
scanf("%d%d",&n,&k);
for(i = ;i <= n;i ++)
scanf("%d",&p[i]);
bin[] = ;
for(i = ;i <= ;i ++)
bin[i] = bin[i-]*;
CL(n);
str = ;
end = n;
while(str < end)
{
mid = (str + end + )/;
if(judge(mid))
str = mid;
else
end = mid - ;
}
printf("%d %d\n",end,judge(end));
for(i = ;i <= n-end+;i ++)
{
if(query(i,i+end-) <= k)
{
printf("%d %d\n",i,i+end-);
}
}
return ;
}

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