牛客寒假算法基础集训营6 题解报告
比赛地址 ： https://ac.nowcoder.com/acm/contest/332#question
A: 考虑一下极限位置就行
#include
using namespace std;
#define ll long long int
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
int main()
{
std::ios::sync_with_stdio(false);
ll n,m, seven,nine;
while(cin >> n >> m)
{
seven = 7*m;
nine = 9 * m;
if(n >9*m || n < 6*m || n < 0)
{
cout << "jgzjgzjgz" << endl;
}
else if(seven <= n)
cout << "0" << endl;
else if(seven > n)
cout << (seven-n) << endl;
}
return 0;
} B ：暴力
#include
using namespace std;
#define ll long long int
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
int main()
{
std::ios::sync_with_stdio(false);
ll n, m, d, x, t, ans;
while(cin >> n >> m >> d >> x)
{
ans = n;
if(ans >= m)
{
cout << "1"<< endl;continue;
}
for(t = 2; t <= x; t++)
{
n += d;
ans += n;
if(ans >= m)
{
cout << t<< endl;break;
}
}
}
return 0;
} C：贪心，从喜爱值最大的开始贪就行
#include
using namespace std;
#define ll long long int
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
struct node
{
int tol, value;
}mp[100005];
bool cmp(node a, node b)
{
return a.value > b.value;
}
int main()
{
std::ios::sync_with_stdio(false);
int n, m, i;
ll ans;
while(cin >> n >> m)
{
ans = 0;
for(i = 0; i < m; i++)
{
cin >> mp[i].tol;
}
for(i = 0; i < m; i++)
{
cin >> mp[i].value;
}
sort(mp,mp+m,cmp);
for(i = 0; i < m && n; i++)
{
if(n >= mp[i].tol)
{
ans += mp[i].tol*mp[i].value;
n -= mp[i].tol;
}
else
{
ans += n * mp[i].value;
n = 0;
break;
}
}
cout << ans << endl;
}
return 0;
}
D：考虑一下 1001 这个坑 然后第一个偶数就全部吃掉，如果奇数就留一口，接下来后面的就先考虑一下前面一个是否留了一口，优先吧前面留的那口吃掉
#include
using namespace std;
#define ll long long int
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
ll mp[100005];
int main()
{
std::ios::sync_with_stdio(false);
int n, i;
ll ans, sum;
while(cin >> n)
{
sum = ans = 0;
for(i = 0; i < n; i++)
{
cin >> mp[i];
}
for(i = 0; i < n; i++)
{
if(mp[i])
{
if(i==0)
{
ans+= mp[i]/2;
mp[i] -= 2*ans;
}
else if(mp[i]%2==0)
{
if(mp[i-1]==0)
{
ans += mp[i]/2;
mp[i]=0;
}
else
{
ans += mp[i]/2;
mp[i]=1;
}
}
else if(mp[i]%2!=0)
{
ans += mp[i]/2;
mp[i] = 1;
if(mp[i-1]==1)
{
mp[i] = 0;
mp[i-1] = 0;
ans++;
}
}
}
}
// cout << sum / 2 << endl;
cout << ans << endl;
}
return 0;
} E：用数组ans[i][j]记住（0,0）到（i，j）有多少城市会被淹，维护一下数组就行
和这道题方法一样： https://blog.csdn.net/lanyanzhiji123asd/article/details/86674537
n*m< 10^6 这个就用动态数组分配一下。
#include
using namespace std;
#define ll long long int
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
//vectormp[1000005];
//vectorvis[1000005];
int *ans[1000005];
int main()
{
int x,y,a,b,q;
std::ios::sync_with_stdio(false);
int n,m,d,i,j,h, cnt;
while(cin >> n >> m >> d)
{
for(i = 1; i <= n; i++)
ans[i] = new int [m+1];
for(i = 1; i <= n; i++)
{
for(j = 1; j <= m; j++)
{
ans[i][j] = 0;
}
}
for(i = 1; i <= n; i++)
{
for(j = 1; j <= m; j++)
{
cin >> h;
ans[i][j] = h>=d;
}
}
for(i = 1;i <= n; i++)
{
for(j = 1; j <= m; j++)
{
if(i-1>0) //cout << "qqq" << endl;
ans[i][j]=ans[i][j]+ans[i-1][j];
if(j-1>0)
ans[i][j]=ans[i][j]+ans[i][j-1];
if(i-1>0&&j-1>0)
ans[i][j]=ans[i][j]-ans[i-1][j-1];
}
}
cin >> q;
while(q--)
{
cin >> a >> b>> x >> y;
cnt = ans[x][y];
if(a-1>0) cnt -= ans[a-1][y];
if(b-1>0) cnt -= ans[x][b-1];
if(a-1&&b-1) cnt += ans[a-1][b-1];
cout << cnt << endl;
}
}
return 0;
} F:最后结束的时候的情况是可以确定的，三种可能 ： 石头剪刀布，由这三种情况反推最开始的可能
#include
using namespace std;
#define ll long long int
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
char str[3]={'R','P','S'};
char ans[10000005];
int mp[3],mp1[3];
void dfs(int num, int l, int r)
{
int i,j;
if(l==r)
{
ans[l] = str[num];
mp1[num]++;
return;
}
int mid = (l+r)/2;
if(num==0)
{
dfs(0,l,mid);
dfs(2,mid+1,r);
}
else if(num==1)
{
dfs(1,l,mid);
dfs(0,mid+1,r);
}
else if(num==2)
{
dfs(1,l,mid);
dfs(2,mid+1,r);
}
int flag = 0;
for(i = l,j = mid+1;i<=mid&&j<=r;i++,j++)
{
if(ans[i]>ans[j])
{
flag = 1;
break;
}
}
if(flag)
{
for(i = l,j = mid+1;i<=mid&&j<=r;i++,j++)
{
swap(ans[i],ans[j]);
}
}
}
int main()
{
std::ios::sync_with_stdio(false);
int r,p,s;
while(cin >> mp[0] >> mp[1]>> mp[2])
{
memset(mp1,0,sizeof(mp1));
int flag = 0,i;
for(i = 0; i <= 2; i++)
{
memset(mp1,0,sizeof(mp1));
dfs(i,0,(mp[0]+mp[1]+mp[2]-1));
// cout << mp1[0] << " " << mp1[1] << " " << mp1[2] << endl;
ans[mp[0]+mp[1]+mp[2]] = '\0';
if(mp1[0]==mp[0]&&mp1[1]==mp[1]&&mp1[2]==mp[2])
{
flag = 1;cout << ans << endl;
break;
}
}
if(!flag)
{
cout << "IMPOSSIBLE" << endl;
}
}
return 0;
}
G:这道题我实在想不明白 放个博客吧，这个博客写的不错 http://keyblog.cn/article-67.html?tdsourcetag=s_pctim_aiomsg
#include
using namespace std;
#define ll long long int
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
ll a,b;
string ans;
ll cnt;
int main()
{
std::ios::sync_with_stdio(false);
int i,j;
while(cin >> a >> b)
{
while(a H：这道题就是求出每头猪从那头猪进化过来花费最少，可以是第1头猪，第二头猪，第三头猪…… 也可以是自己，所以用 i 枚举表示进化多少次，求出最小值就行 这个方法真是妙呀 我是一个小菜鸡小菜鸡o(╥﹏╥)o
#include
using namespace std;
#define ll long long int
const ll INF = 0x3f3f3f3f3f3f3f3f;//这里注意一下，不然会WA
const int mod = 998244353;
ll b[20005],mp[20005],x;
int main()
{
std::ios::sync_with_stdio(false);
int n, i, j;
ll ans;
while(cin >> n >> x)
{
ans = INF;
memset(b,INF, sizeof(b));
for(i = 0; i < n; i++)
cin >> mp[i];
for(i = 0; i < n; i++)
{
ll sum = 0;
for(j = 0; j < n; j++)
b[j] = min(b[j],mp[(j-i+n)%n]);
for(j = 0; j < n; j++)
sum += b[j];
ans = min(ans, sum+i*x);
}
cout <
I ： 一串01组成的序列，如果00 或11配对了就直接算10，用栈维护下，这样最后一定剩下的是01交叠的串，直接算就行
#include
using namespace std;
#define ll long long int
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
string str;
int mp[1000005];
int main()
{
std::ios::sync_with_stdio(false);
int ans,cnt, i;
while(cin >> str)
{
memset(mp,-1,sizeof(mp));
ans = cnt = 0;
mp[cnt++] = str[0]-'0';
for(i = 1; i < str.length(); i++)
{
if(mp[cnt-1]!=str[i]-'0' || cnt==0)
{
mp[cnt++] = str[i]-'0';
}
else
{
cnt--;
ans += 10;
}
}
ans += 5*(cnt/2);
cout << ans << endl;
}
return 0;
} J ： 简单BFS 记住一下当前左边右边走了多少步判断一下就行
#include
using namespace std;
#define ll long long int
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
int n,m,r,c,x,y;
int d[2] = {1,-1};
struct node
{
int x,y,left,right;
};
int vis[1005][1005];
string str[1005];
int access(node tt)
{
if(tt.x=0&&tt.y=0&&!vis[tt.x][tt.y]&&str[tt.x][tt.y]!='*')
return 1;
return 0;
}
void BFS()
{
memset(vis, 0, sizeof(vis));
int ans = 0, i;
node temp;
temp.x=r;
temp.y=c;
temp.left = 0;
temp.right=0;
queueq;
q.push(temp);
vis[temp.x][temp.y] = 1;
ans++;
while(!q.empty())
{//cout<< "qqq" << endl;
node v = q.front();
q.pop();
temp.x = v.x;
temp.y = v.y-1;
temp.left = v.left+1;
temp.right = v.right;
if(access(temp))
{
if(temp.left<=x)
{
q.push(temp);
vis[temp.x][temp.y] = 1;
ans++;
}
}
temp.x = v.x;
temp.y = v.y+1;
temp.left = v.left;
temp.right = v.right+1;
if(access(temp))
{
if(temp.right<=y)
{
q.push(temp);
vis[temp.x][temp.y] = 1;
ans++;
}
}
for(i = 0; i <= 1; i++)
{
temp.x = v.x+d[i];
temp.y = v.y;
temp.left = v.left;
temp.right = v.right;
if(access(temp))
{
q.push(temp);
vis[temp.x][temp.y] = 1;
ans++;
}
}
}
cout << ans << endl;
}
int main()
{
std::ios::sync_with_stdio(false);
int i;
while(cin >> n >> m)
{
cin >> r >> c;
cin >> x>> y;
r--,c--;
for(i = 0; i < n; i++)
{
cin >> str[i];
}
BFS();
}
return 0;
}