A:
简单的想法题。使用hash表或者是map存储所有数，然后从最小的数开始找从这个数开始的连续P倍数的个数X，那么需要删除的数的个数为X/2。
代码如下：
Problem A
B:
组合数学题，通过经典的一一对应原则推导出答案为N^(N-2)。http://wenku.baidu.com/view/2e1ab2757fd5360cba1adbba.html
Cayley定理在组合数学中的应用。代码如下：
Problem B #include#include #include #include #include #include using namespace std;typedef long long LL;const LL M=1e9+7;LL Pow( LL a, LL b){ LL ans=1; while( b ){ if( b&1 ){ ans*=a; ans%=M; } a*=a; a%=M; b>>=1; } return ans;} int main( ){ LL N; while( scanf( "%lld", &N )!=EOF ){ printf( "%lld\n", Pow( N, N-2 ) ); } return 0;}
C:
计算几何。该题一个简单的做法是先找出三个点的外心，然后用极坐标每次旋转72°生成剩下的点。
代码如下：
Problem C #include#include #include #include #include #include #include using namespace std;const double PI = acos(-1);struct point{ double x,y;};struct line{point a,b;};vector v;point p[3], cc;point intersection(line u,line v){ point ret=u.a; double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x)) /((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x)); ret.x+=(u.b.x-u.a.x)*t; ret.y+=(u.b.y-u.a.y)*t; return ret;}point circumcenter(point a,point b,point c){ line u,v; u.a.x=(a.x+b.x)/2; u.a.y=(a.y+b.y)/2; u.b.x=u.a.x-a.y+b.y; u.b.y=u.a.y+a.x-b.x; v.a.x=(a.x+c.x)/2; v.a.y=(a.y+c.y)/2; v.b.x=v.a.x-a.y+c.y; v.b.y=v.a.y+a.x-c.x; return intersection(u,v);}double dist(point a, point b) { return sqrt(double( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) )); }bool cmp(point a, point b) { if (fabs(a.x - b.x) > 1e-6) { return a.x < b.x; } else { return a.y < b.y; }}int main() { double R; point info; while (scanf("%lf %lf", &p[0].x, &p[0].y) != EOF) { v.clear(); for (int i = 1; i < 3; ++i) { scanf("%lf %lf", &p[i].x, &p[i].y); } cc = circumcenter(p[0], p[1], p[2]); R = dist(p[0], cc); double phi = atan2(p[0].y-cc.y, p[0].x-cc.x); // 求出第一个点的角度 for (int i = 0; i < 5; ++i) { info.x = R * cos(phi) + cc.x, info.y = R * sin(phi) + cc.y; v.push_back(info); phi += 72.0*PI/180.0; } sort(v.begin(), v.end(), cmp); for (int i = 0; i < 5; ++i) { int flag = 0; for (int j = 0; j < 3; ++j) { if (fabs(v[i].x - p[j].x) < 1e-6 && fabs(v[i].y - p[j].y) < 1e-6) { flag = 1; break; } } if (!flag) { printf("%.2f %.2f\n", v[i].x, v[i].y); } } } return 0; }
D:
动态规划。题意是求将一个串改造成回文串的最少花费。给出所有组成串的字母添加和删除的花费。设dp[i][j]表示[i,j]子串被改造成回文串的最少开销是多少。那么有动态规划方程：
dp[i][j] = min(dp[i+1][j] + cost[i], dp[i][j-1] + cost[j]); 其中cost[i] = min(add[i], del[i])存储删除和添加节点的较小的花费。如果str[i] == str[j]的话就多出一种选择dp[i-1][j-1]。代码如下：
Problem D #include#include #include #include #include using namespace std;int N, M, dp[2005][2005];int cost[30];char str[2005];int DP() { for (int i = 1; i < M; ++i) { for (int j = 1; j+i <= M; ++j) { int a = j, b = j + i, ll = str[a]-'a', rr = str[b]-'a'; dp[a][b] = min(dp[a+1][b] + cost[ll], dp[a][b-1] + cost[rr]); if (ll == rr) dp[a][b] = min(dp[a][b], dp[a+1][b-1]); } } return dp[1][M];}int main() { char s[5]; int a, b; while (scanf("%d %d", &N, &M) == 2) { scanf("%s", str+1); for (int i = 1; i <= N; ++i) { scanf("%s %d %d", s, &a, &b); cost[s[0]-'a'] = min(a, b); } printf("%d\n", DP()); } return 0;}
E:
分解素因子，转换成M个球放到N个盒子模型，得出组合公式C(N+M-1, N-1),因为ai>1,既盒子不允许为空，所以求出总方案数后，减去盒子为空的情形。对于盒子为空的求解，通过假设至少一个盒子（C(N+M-1-1,M-1-1)），两个，直到N-1个，由容斥来解决重复问题。注意单个素因子，其次数可能会很大，所以C预处理要比较大，不然会RE。
代码如下：
Problem E #include#include #include #include #include #include
F:
并查集+最短路。通过并查集处理结盟的点，然后从集合中选择一个到目标点最短的距离输出。
代码如下：
Problem F #include#include #include using namespace std;const int inf = 0x7f7f7f7f;const int MAXV = 1002;int UFS[MAXV];int Find(int x){ return UFS[x]=(UFS[x]!=x?Find(UFS[x]):x);}void Union(int x,int y){ UFS[Find(x)]=Find(y);}char o;int n,m,k,u,v,w,x,y,a[MAXV][MAXV];int main(){ while(scanf("%d %d %d",&n,&m,&k)!=EOF) { memset(a,inf,sizeof(a)); for(int i=0;n>i;i++) { UFS[i]=i; } for(int i=0;m>i;i++) { scanf("%d %d %d",&u,&v,&w); a[u][v]=min(a[u][v],w); } for(int z=0;n>z;z++) { for(int i=0;n>i;i++) { if(a[i][z]!=inf) { for(int j=0;n>j;j++) { if(a[z][j]!=inf) { a[i][j]=min(a[i][j],a[i][z]+a[z][j]); } } } } } for(int i=0;k>i;i++) { getchar(); scanf("%c %d %d",&o,&x,&y); if(o=='I') Union(x,y); else { int ans=inf; for(int i=0;n>i;i++) { if(Find(x)==Find(i)) { ans=min(ans,a[i][y]); } } printf("%d\n",ans!=inf?ans:-1); } } } return 0;}
G:
二分枚举构图+二分匹配。该题想说明的就是A中的一个城市只能够对应B中的一个城市。先通过floyd求出任意城市之间的最短路，然后二分枚举构好A->B城市的二分图，然后使用匈牙利算法计算是否能够达到完全匹配。
代码如下：
Problem G #include#include #include #include #include #include using namespace std;const int MAX = 105;const int INF = 0x3f3f3f3f;int n, N, M, reca[MAX], recb[MAX];int mp[MAX][MAX];int marry[MAX];char G[MAX][MAX];char vis[MAX];void floyd() { for (int k = 0; k < n; ++k) { for (int i = 0; i < n; ++i) { if (mp[i][k] == INF || i == k) continue; for (int j = 0; j < n; ++j) { if (mp[k][j] == INF || j == k) continue; if (mp[i][j] > mp[i][k] + mp[k][j]) { mp[i][j] = mp[i][k] + mp[k][j]; } } } }}bool path(int u) { for (int i = 0; i < M; ++i) { if (!G[u][i] || vis[i]) continue; vis[i] = 1; if (marry[i] == -1 || path(marry[i])) { marry[i] = u; return true; } } return false;}void build(int lim) { memset(G, 0, sizeof (G)); for (int i = 0; i < N; ++i) { for (int j = 0; j < M; ++j) { if (mp[reca[i]][recb[j]] <= lim) { G[i][j] = 1; } } }}bool Ac() { int cnt = 0; memset(marry, 0xff, sizeof (marry)); for (int i = 0; i < N; ++i) { memset(vis, 0, sizeof (vis)); if (path(i)) { ++cnt; } } return cnt == M;}int bsearch(int l, int r) { int mid, ret; while (l <= r) { mid = (l + r) >> 1; build(mid); if (Ac()) { ret = mid; r = mid - 1; } else { l = mid + 1; } } return ret;}int main(){ while (scanf("%d", &n) == 1) { for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { scanf("%d", &mp[i][j]); } } scanf("%d", &N); for (int i = 0; i < N; ++i) { scanf("%d", &reca[i]); --reca[i]; } scanf("%d", &M); for (int i = 0; i < M; ++i) { scanf("%d", &recb[i]); --recb[i]; } floyd(); printf("%d\n", bsearch(0, 10000)); } return 0;}
H:
法一：通过题目给定的关系不难知道a[i] = 3*a[i-1] + a[i-2] - 3*a[i-3]。假设要求第D天的口令，那么计算出a[D-1], a[D-2], a[D-3]就能够计算出a[D]，求出四个系数后，代入X便可求解答案了。这题D给的数据比较大，直接计算肯定会TLE，那么使用矩阵并加上快速降幂来计算就简单多了，将给定的b,c,d作为3*1的矩阵B{a[0], a[-1], a[-2]}^-1，矩阵A=
3 1 -31 0 00 1 0作为递推矩阵，那么在B矩阵左边乘上D-1个A最后的结果就是{a[D-1], a[D-2], a[D-3]}^-1，D-1个A相乘就可以使用矩阵快速幂了。代码如下：
Problem H #include#include #include #include #include #include using namespace std;typedef long long int64;const int MOD = int(1e9)+7;int a, b, c, d, D, X;int64 R[3][3] = { { 3, 1, -3}, { 1, 0, 0}, { 0, 1, 0}};struct Matrix { int64 mat[3][3]; Matrix() { memset(mat, 0, sizeof (mat)); } void unit() { memset(mat, 0, sizeof (mat)); for (int i = 0; i < 3; ++i) { mat[i][i] = 1; } } void init() { memcpy(mat, R, sizeof (mat)); } void show() const; friend Matrix operator * (const Matrix & a, const Matrix & b); friend Matrix pow(Matrix a, int b);};void Matrix::show() const { for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { printf("%d ", mat[i][j]); } puts(""); }}Matrix operator * (const Matrix & a, const Matrix & b) { Matrix ret; for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { for (int k = 0; k < 3; ++k) { ret.mat[i][j] += (a.mat[i][k] * b.mat[k][j]) % MOD; ret.mat[i][j] %= MOD; } } } return ret;}Matrix pow(Matrix a, int b) { Matrix ret; ret.unit(); while (b) { if (b & 1) { ret = ret * a; } a = a * a; b >>= 1; } return ret;}int64 cal(int64 na, int64 nb, int64 nc, int64 nd) { int64 ini[4] = { 1}; for (int i = 1; i < 4; ++i) { ini[i] = (ini[i-1] * X) % MOD; } return ((na*ini[3])%MOD+(nb*ini[2])%MOD+(nc*ini[1])%MOD+(nd*ini[0])%MOD)%MOD;}void solve() { Matrix mm; mm.init(); mm = pow(mm, D-1); int64 na, nb, nc, nd; nb = ((mm.mat[0][0]*b)%MOD + (mm.mat[0][1]*c)%MOD + (mm.mat[0][2]*d)%MOD)%MOD; nc = ((mm.mat[1][0]*b)%MOD + (mm.mat[1][1]*c)%MOD + (mm.mat[1][2]*d)%MOD)%MOD; nd = ((mm.mat[2][0]*b)%MOD + (mm.mat[2][1]*c)%MOD + (mm.mat[2][2]*d)%MOD)%MOD; na = (3*nb)%MOD+nc-(3*nd)%MOD; printf("%d\n", int((cal(na, nb, nc, nd)+MOD)%MOD));}int main() { while (scanf("%d %d %d %d", &a, &b, &c, &d) != EOF) { assert(a >= 0 && a <= 100); scanf("%d %d", &D, &X); solve(); } return 0;}
法二：根据表达式a[i] = 3*a[i-1] + a[i-2] - 3*a[i-3]，由于该表达式为常系数3阶齐次递推关系，因此设a[n] = r^n，代入方程求出特征根为-1,1,3，因此a[n]=A*(-1)^n+B*(1)^n+C*(3)^n，令a[0]=d,a[1]=c,a[2]=b求出A,B,C之后再根据通项公式求出D天后的系数。b,c,d是8的倍数保证了解三元一次方程组时不会出现分数。
代码如下：
Problem H #include#include #include #include using namespace std;typedef long long int64;const int64 MOD = (int64)1e9 + 7;int64 a, b, c, d, D, X;int64 _pow(int64 a, int64 b) { int64 ret = 1; while (b) { if (b & 1) { ret *= a; ret %= MOD; } b >>= 1; a *= a; a %= MOD; } return ret;}int64 cal(int64 na, int64 nb, int64 nc, int64 nd) { int64 ini[4] = { 1}; for (int i = 1; i < 4; ++i) { ini[i] = (ini[i-1] * X) % MOD; } return ((na*ini[3])%MOD+(nb*ini[2])%MOD+(nc*ini[1])%MOD+(nd*ini[0])%MOD)%MOD;}int main() { while (scanf("%lld %lld %lld %lld", &a, &b, &c, &d) != EOF) { scanf("%lld %lld", &D, &X); int64 A = (3*d-4*c+b)/8; int64 B = (3*d+2*c-b)/4; int64 C = (b-d)/8; a = (A * ((D+2) & 1 ? -1 : 1) + B + C * _pow(3, D+2)) % MOD; b = (A * ((D+1) & 1 ? -1 : 1) + B + C * _pow(3, D+1)) % MOD; c = (A * ((D) & 1 ? -1 : 1) + B + C * _pow(3, D)) % MOD; d = (A * ((D-1) & 1 ? -1 : 1) + B + C * _pow(3, D-1)) % MOD; printf("%lld\n", (cal(a, b, c, d)+MOD)%MOD); } return 0; }