BestCoder Round #56 1002 Clarke and problem 1003 Clarke and puzzle （dp，二维bit或线段树）
今天第二次做BC，不习惯hdu的oj，CE过2次。。。
1002 Clarke and problem
和Codeforces Round #319 (Div. 2) B Modulo Sum思路差不多，
将a[i]对p取余数，最后得到0的方案总数即使答案，dp转移，一个状态方案总数等于能转移过来的状态方案数之和
#include<cstdio> #include<iostream> #include<string> #include<cstring> #include<queue> #include<vector> #include<stack> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; #define PB push_back #define MP make_pair #define fi first #define se second #define cer(x) cout<<#x<<'='<<endl typedef long long ll; const int maxn = 1050; int dp[2][maxn]; //#define LOCAL const int MOD = 1e9+7; int main() { #ifdef LOCAL freopen("in.txt","r",stdin); #endif int T; scanf("%d",&T); while(T--){ int n,p; scanf("%d%d",&n,&p); //fill(dp[0],dp[0]+p,0); memset(dp[0],0,sizeof(dp[0])); dp[0][0] = 1; for(int i = 0; i < n; i++){ int c = i&1,nx = (i&1)^1; //fill(dp[nx],dp[nx]+p,0); memcpy(dp[nx],dp[c],sizeof(dp[nx])); int a; scanf("%d",&a); a %= p; if(a<0) a += p; //G++编译器负数取模还是负数，在这里RE了几次 for(int j = 0; j < p; j++){ if(dp[c][j]){ int t = (j+a)%p; dp[nx][t] = (dp[nx][t]+ dp[c][j])%MOD; } } } printf("%d\n",dp[n&1][0]); } return 0; }
1003 Clarke and puzzle
熟悉NIM游戏的结论和SG函数思路上并不难。
卡时间卡的太紧了，按照官方题解做法980ms过。。。把memset换成for，890ms
自信地写了个二维线段树被卡。写BIT还遇到一些奇怪的错误。
#include<cstdio> #include<iostream> #include<string> #include<cstring> #include<queue> #include<vector> #include<stack> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; #define PB push_back #define MP make_pair #define fi first #define se second #define cer(x) cout<<#x<<'='<<endl typedef long long ll; const int maxn = 505; int C[maxn][maxn],n,m,c[maxn][maxn]; #define lb(x) ((x)&-(x)) int sum(int x,int y){ int r = 0; for(int i=x;i>0;i-=lb(i)) //for(; x>0; x-=lb(x)) 奇怪的错误，写成这样T了 for(int j=y;j>0;j-=lb(j)) r ^= C[i][j]; return r; } void add(int x,int y,int val){ for(int i=x;i<=n;i+=lb(i)) for(int j=y;j<=m;j+=lb(j)) C[i][j] ^= val; } //#define LOCAL int main() { #ifdef LOCAL freopen("in.txt","r",stdin); #endif int T; scanf("%d",&T); while(T--){ int q; scanf("%d%d%d",&n,&m,&q); //memset(C,0,sizeof(C)); for(int i = 1; i <= n; i++){ //fill(C[i]+1,C[i]+1+m,0); for(int j = 1; j <= m; j++) C[i][j] = 0; } for(int i = 1; i <= n; i++){ for(int j = 1; j <= m; j++){ scanf("%d",c[i]+j); add(i,j,c[i][j]); } } for(int i = 0; i < q; i++){ int op; scanf("%d",&op); if(op == 1){ int x1,y1,x2,y2; scanf("%d%d%d%d",&x1,&y1,&x2,&y2); puts(sum(x2,y2)^sum(x2,y1-1)^sum(x1-1,y2)^sum(x1-1,y1-1)?"Yes":"No"); }else { int x,y,v; scanf("%d%d%d",&x,&y,&v); add(x,y,c[x][y]^v); c[x][y] = v; } } } return 0; }
重新改了时限后二维线段树过了
#include<cstdio> #include<iostream> #include<string> #include<cstring> #include<queue> #include<vector> #include<stack> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; #define PB push_back #define MP make_pair #define fi first #define se second #define cer(x) cout<<#x<<'='<<endl typedef long long ll; const int maxn = 502<<2,N = 502; int sx[maxn][maxn],n,m; int c[N][N]; int xo,xleaf,x1,y1,x2,y2,x,y,v,vsx; #define lid (id<<1) #define rid (id<<1|1) void query1D(int id = 1,int l = 1,int r = m) { if(y1 <= l && r <= y2){ vsx ^= sx[xo][id]; }else { int mid = (l+r)>>1; if(y1 <= mid) query1D(lid,l,mid); if(mid < y2) query1D(rid,mid+1,r); } } void query2D(int id = 1,int l = 1,int r = n) { if(x1 <= l && r <= x2) { xo = id; query1D(); } else { int mid = (l+r)>>1; if(x1 <= mid) query2D(lid,l,mid); if(mid < x2) query2D(rid,mid+1,r); } } void modify1D(int id = 1, int l = 1, int r = m) { if(l == r){ if(xleaf) { sx[xo][id] = v; return; } sx[xo][id] = sx[xo<<1][id]^sx[xo<<1|1][id]; }else { int mid = (l+r)>>1; if(y <= mid) modify1D(lid,l,mid); else modify1D(rid,mid+1,r); sx[xo][id] = sx[xo][lid]^sx[xo][rid]; } } void modify2D(int id = 1, int l = 1, int r = n) { if(l==r) { xo = id; xleaf = 1; modify1D(); } else { int mid = (l+r)>>1; if(x<=mid) modify2D(lid,l,mid); else modify2D(rid,mid+1,r); xo = id; xleaf = 0; modify1D(); } } void build1D(int id = 1,int l = 1,int r = m) { if(l == r) { if(xleaf) { sx[xo][id] = c[x][l]; return; } sx[xo][id] = sx[xo<<1][id]^sx[xo<<1|1][id]; } else { int mid = (l+r)>>1,lc = lid,rc = rid; build1D(lc,l,mid); build1D(rc,mid+1,r); sx[xo][id] = sx[xo][lc]^sx[xo][rc]; } } void build2D(int id = 1, int l = 1, int r = n) { if(l == r){ xo = id; xleaf = 1; x = l; build1D(); } else { int mid = (l+r)>>1; build2D(lid,l,mid); build2D(rid,mid+1,r); xo = id; xleaf = 0; build1D(); } } inline int read() { char c; while((c=getchar())<'0'||c>'9'); int ret=c-'0'; while((c = getchar())>='0'&&c<='9') ret = ret*10+(c-'0'); return ret; } //#define LOCAL int main() { #ifdef LOCAL freopen("in.txt","r",stdin); #endif int T; scanf("%d",&T); while(T--){ n = read(); m = read(); int q = read(); for(int i = 1; i <= n; i++){ for(int j = 1; j <= m; j++){ c[i][j] = read(); } } build2D(); while(q--){ if(read() == 1){ x1 = read(); y1= read(); x2 = read(); y2 = read(); vsx = 0; query2D(); puts(vsx?"Yes":"No"); }else { x = read(); y = read(); v = read(); modify2D(); } } } return 0; }
