Codeforces Round #427 (Div. 2) C D

C

对坐标维护一个前缀和
sum[i][j]表示以[1,1]为左上角 [i,j]为右下角的数量
每次查询一个矩形只要剪掉上边和左边多余的矩形再加回多减的部分就可以了
因为还有亮度 再增加一维记录每个亮度的星星情况

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define slld(n) scanf("%lld",&n)#define slldd(n,m) scanf("%lld%lld",&n,&m)#define sllddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define slf(n) scanf("%lf",&n)#define slff(n,m) scanf("%lf%lf",&n,&m)#define slfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define llans() printf("%lld",ans)#define llanss() printf("%lld ",ans)#define llansn() printf("%lld\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<v.size();++i)#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define mp make_pair#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))typedef long long ll;typedef unsigned long long ull;const int mod = 1000000000+7;const double eps=1e-7;const int inf=0x3f3f3f3f;const ll infl = 1e16;const int maxn=  200+10;const int maxm = 5000000+10;int sum[maxn][maxn][maxn];int a[maxn][maxn][maxn];int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    int n,q,c;    sddd(n,q,c);    for(int i=1; i<=n; ++i)    {        int x,y,s;        sddd(x,y,s);        a[s][x][y]++;    }    for(int k=0; k<=10; ++k)        for(int i=1; i<=100; ++i)        {            for(int j=1; j<=100; ++j)            {                sum[k][i][j] = sum[k][i][j-1]+a[k][i][j];            }        }    for(int k=0; k<=10; ++k)        for(int i=1; i<=100; ++i)        {            for(int j=1; j<=100; ++j)            {                sum[k][i][j] = sum[k][i-1][j]+sum[k][i][j];            }        }    for(int i=1; i<=q; ++i)    {        int t;        sd(t);        int x1,y1,x2,y2;        sdd(x1,y1),sdd(x2,y2);        int ans = 0;        for(int k=0;k<=10;++k)        {            int next = (k+t)%(c+1);            int idx = sum[k][x2][y2] - sum[k][x2][y1-1] - sum[k][x1-1][y2] + sum[k][x1-1][y1-1];            ans += next *idx;        }        ansn();    }    return 0;}

D

按题意模拟一下
先n^2处理每个子串是否为回文串
再处理出每个子串的回文度

如果一段是回文串 它的两半是回文串 那两半一定是相等的
实际上 回文串的左右两半的子串的回文度一定是相等的

假如一个子串的最高回文度是k
那它同时也是回文度为k-1,k-2,……,1的串

再维护一个后缀和

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define sld(n) scanf("%lld",&n)#define sldd(n,m) scanf("%lld%lld",&n,&m)#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define sf(n) scanf("%lf",&n)#define sff(n,m) scanf("%lf%lf",&n,&m)#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define lans() printf("%lld",ans)#define lanss() printf("%lld ",ans)#define lansn() printf("%lld\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<(int)v.size();++i)#define szz(x) ((int)x.size())#define ms(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define mp make_pair#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))typedef long long ll;typedef unsigned long long ull;const ll mod = 1000000000+7;const double eps=1e-7;const int inf=0x3f3f3f3f;const ll infl = 1e16;const int maxn=  5000+10;const int maxm = 1000000+10;char s[maxn];int p[maxn][maxn];int dp[maxn][maxn];int cnt[maxn];int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    ss(s+1);    int n = strlen(s+1);    r1(i,n)p[i][i] = p[i][i-1] = 1;    for(int i=n;i;--i)    {        for(int j= i+1;j<=n;++j)            if(s[i]==s[j]&&p[i+1][j-1])p[i][j]=1;    }    for(int i=1;i<=n;++i)dp[i][i]=1;    for(int i=n;i;--i)    {        for(int j = i+1;j<=n;++j)        {            if(!p[i][j])continue;            int len = j-i+1;            dp[i][j] = min(dp[i][i+len/2-1],dp[j-len/2+1][j])+1;        }    }    for(int i=1;i<=n;++i)    {        for(int j=1;j<=n;++j)            ++cnt [ dp[i][j] ];    }    for(int i=n-1;i;--i)cnt[i] +=cnt[i+1];    r1(i,n)printf("%d ",cnt[i]);    return 0;}