搞了好久才把大部分题目题解看完了，真是太弱了。
A题简单暴力题一个一个匹配，对应位置字母要么相同，要么是'.'.
B题给定一个矩阵，左下角(0,0)，右上角(n, m)，取4个不同的点连成一段折线，要有最长的折线长度。
排除n == 0 和m == 0 ，剩下的情况中总共由4中情况：
枚举一下就可以了
1. (0,0)->(n,m)->(0, m)->(n, 0)
2. (0,0)->(n,m)->(n, 0)->(0, m)
3.(n,m-1)->(0,0)->(n,m)->(0,1)
4.(0,m-1)->(n,0)->(0,m)->(n,1)
代码：
1 //Template updates date: 20140718 2 #include <iostream> 3 #include <sstream> 4 #include <cstdio> 5 #include <climits> 6 #include <ctime> 7 #include <cctype> 8 #include <cstring> 9 #include <cstdlib> 10 #include <string> 11 #include <stack> 12 #include <set> 13 #include <map> 14 #include <cmath> 15 #include <vector> 16 #include <queue> 17 #include <algorithm> 18 #define esp 1e-6 19 #define inf 0x3f3f3f3f 20 #define pi acos(-1.0) 21 #define pb push_back 22 #define lson l, m, rt<<1 23 #define rson m+1, r, rt<<1|1 24 #define lowbit(x) (x&(-x)) 25 #define mp(a, b) make_pair((a), (b)) 26 #define bit(k) (1<<(k)) 27 #define iin freopen("pow.in", "r", stdin); 28 #define oout freopen("pow.out", "w", stdout); 29 #define in freopen("solve_in.txt", "r", stdin); 30 #define out freopen("solve_out.txt", "w", stdout); 31 #define bug puts("********))))))"); 32 #define Inout iin oout 33 #define inout in out 34 35 #define SET(a, v) memset(a, (v), sizeof(a)) 36 #define SORT(a) sort((a).begin(), (a).end()) 37 #define REV(a) reverse((a).begin(), (a).end()) 38 #define READ(a, n) {REP(i, n) cin>>(a)[i];} 39 #define REP(i, n) for(int i = 0; i < (n); i++) 40 #define VREP(i, n, base) for(int i = (n); i >= (base); i--) 41 #define Rep(i, base, n) for(int i = (base); i < (n); i++) 42 #define REPS(s, i) for(int i = 0; (s)[i]; i++) 43 #define pf(x) ((x)*(x)) 44 #define mod(n) ((n)) 45 #define Log(a, b) (log((double)b)/log((double)a)) 46 #define Srand() srand((int)time(0)) 47 #define random(number) (rand()%number) 48 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a) 49 50 using namespace std; 51 typedef long long LL; 52 typedef unsigned long long ULL; 53 typedef vector<int> VI; 54 typedef pair<int,int> PII; 55 typedef vector<PII> VII; 56 typedef vector<PII, int> VIII; 57 typedef VI:: iterator IT; 58 typedef map<string, int> Mps; 59 typedef map<int, int> Mpi; 60 typedef map<int, PII> Mpii; 61 typedef map<PII, int> Mpiii; 62 double ans[4]; 63 64 int main() { 65 66 int n, m; 67 cin>>n>>m; 68 if(n == 0) { 69 printf("%d %d ", 0, 1); 70 printf("%d %d ", 0, m); 71 printf("%d %d ", 0, 0); 72 printf("%d %d ", 0, m-1); 73 } else if(m == 0) { 74 printf("%d %d ", 1, 0); 75 printf("%d %d ", n, 0); 76 printf("%d %d ", 0, 0); 77 printf("%d %d ", n-1, 0); 78 } else { 79 ans[0] = m + 2*sqrt(pf(n)+pf(m)); 80 ans[1] = n + 2*sqrt(pf(n)+pf(m)); 81 ans[2] = sqrt(pf(n)+pf(m))+sqrt(pf(n)+pf(m-1))*2; 82 ans[3] = sqrt(pf(n)+pf(m))+sqrt(pf(n-1)+pf(m))*2; 83 int tmp = 0; 84 REP(i, 4) 85 if(ans[tmp] < ans[i]) 86 tmp = i; 87 if(tmp == 0) { 88 printf("%d %d ", 0, 0); 89 printf("%d %d ", n, m); 90 printf("%d %d ", n, 0); 91 printf("%d %d ", 0, m); 92 } else if(tmp == 1) { 93 printf("%d %d ", 0, 0); 94 printf("%d %d ", n, m); 95 printf("%d %d ", 0, m); 96 printf("%d %d ", n, 0); 97 } else if(tmp == 3) { 98 printf("%d %d ", 1, m); 99 printf("%d %d ", n, 0); 100 printf("%d %d ", 0, m); 101 printf("%d %d ", n-1, 0); 102 } else { 103 printf("%d %d ", n, m-1); 104 printf("%d %d ", 0, 0); 105 printf("%d %d ", n, m); 106 printf("%d %d ", 0, 1); 107 } 108 } 109 return 0; 110 }
C题:有n种卡片，每种m张,魔术师每次从中选取n张然后给观众选取，自己再从中选出一张，问最后选得的和观众相同的概率是多少？
分析：由于最后选得的相同的卡片肯定是n种中一种，而且是哪种不重要，这样只需考虑，该种卡片在选出的n张牌占多少？
选出卡片中该种有i张，然后最后魔术师和观众选中该种卡片，这样概率就是：C(m,i)*C((n-1)m, n-i)*(i/n)^2/C(nm,n) 1<=i<=min(n, m)
由于对应n种卡片都是上面这样计算的，所以上面结果*n就是最后结果了。
最后计算时候可以用log这样乘除法转化为+，-,对于概率计算很方便，对结果取一下e^ans放就可以了。
1 //Template updates date: 20140718 2 #include <iostream> 3 #include <sstream> 4 #include <cstdio> 5 #include <climits> 6 #include <ctime> 7 #include <cctype> 8 #include <cstring> 9 #include <cstdlib> 10 #include <string> 11 #include <stack> 12 #include <set> 13 #include <map> 14 #include <cmath> 15 #include <vector> 16 #include <queue> 17 #include <algorithm> 18 #define esp 1e-6 19 #define inf 0x3f3f3f3f 20 #define pi acos(-1.0) 21 #define pb push_back 22 #define lson l, m, rt<<1 23 #define rson m+1, r, rt<<1|1 24 #define lowbit(x) (x&(-x)) 25 #define mp(a, b) make_pair((a), (b)) 26 #define bit(k) (1<<(k)) 27 #define iin freopen("pow.in", "r", stdin); 28 #define oout freopen("pow.out", "w", stdout); 29 #define in freopen("solve_in.txt", "r", stdin); 30 #define out freopen("solve_out.txt", "w", stdout); 31 #define bug puts("********))))))"); 32 #define Inout iin oout 33 #define inout in out 34 35 #define SET(a, v) memset(a, (v), sizeof(a)) 36 #define SORT(a) sort((a).begin(), (a).end()) 37 #define REV(a) reverse((a).begin(), (a).end()) 38 #define READ(a, n) {REP(i, n) cin>>(a)[i];} 39 #define REP(i, n) for(int i = 0; i < (n); i++) 40 #define VREP(i, n, base) for(int i = (n); i >= (base); i--) 41 #define Rep(i, base, n) for(int i = (base); i < (n); i++) 42 #define REPS(s, i) for(int i = 0; (s)[i]; i++) 43 #define pf(x) ((x)*(x)) 44 #define mod(n) ((n)) 45 #define Log(a, b) (log((double)b)/log((double)a)) 46 #define Srand() srand((int)time(0)) 47 #define random(number) (rand()%number) 48 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a) 49 50 using namespace std; 51 typedef long long LL; 52 typedef unsigned long long ULL; 53 typedef vector<int> VI; 54 typedef pair<int,int> PII; 55 typedef vector<PII> VII; 56 typedef vector<PII, int> VIII; 57 typedef VI:: iterator IT; 58 typedef map<string, int> Mps; 59 typedef map<int, int> Mpi; 60 typedef map<int, PII> Mpii; 61 typedef map<PII, int> Mpiii; 62 double nCr(int n, int m) { 63 double res = 0; 64 Rep(i, 1, m+1) 65 res += -log(i)+log(n-i+1); 66 return res; 67 } 68 void solve(int n, int m) { 69 int M = min(n, m); 70 double ans = 0; 71 Rep(i, 1, M+1) { 72 double tmp = 2*(log(i)-log(n))+log(n)+nCr(m, i)+nCr((n-1)*m, n-i)-nCr(n*m, n); 73 ans += exp(tmp); 74 } 75 printf("%.9f ", ans); 76 } 77 int main() { 78 int n, m; 79 cin>>n>>m; 80 solve(n, m); 81 return 0; 82 }
D题：有k件衣服，每件必须先后并且连续经过3种操作，洗涤，烘干，折叠，分别由3种机器来完成，其中机器数目为n1,n2,n3,一个机器只能同时对一件衣服进行对应操作。其中衣服完成每种操作的时间为t1,t2,t3.求最后完成所有衣服3种操作的最短时间。
分析：由于每种操作最多可能处理ni件衣服所以第ni+1件衣服和第1件衣服的开始处理时间必须相差ti，根据这个可以推理最终完成时间。
代码：
//Template updates date: 20140718 #include <bits/stdc++.h> #define esp 1e-6 #define inf 0x3f3f3f3f #define pi acos(-1.0) #define pb push_back #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 #define lowbit(x) (x&(-x)) #define mp(a, b) make_pair((a), (b)) #define bit(k) (1<<(k)) #define iin freopen("pow.in", "r", stdin); #define oout freopen("pow.out", "w", stdout); #define in freopen("solve_in.txt", "r", stdin); #define out freopen("solve_out.txt", "w", stdout); #define bug puts("********))))))"); #define Inout iin oout #define inout in out #define SET(a, v) memset(a, (v), sizeof(a)) #define SORT(a) sort((a).begin(), (a).end()) #define REV(a) reverse((a).begin(), (a).end()) #define READ(a, n) {REP(i, n) cin>>(a)[i];} #define REP(i, n) for(int i = 0; i < (n); i++) #define VREP(i, n, base) for(int i = (n); i >= (base); i--) #define Rep(i, base, n) for(int i = (base); i < (n); i++) #define REPS(s, i) for(int i = 0; (s)[i]; i++) #define pf(x) ((x)*(x)) #define mod(n) ((n)) #define Log(a, b) (log((double)b)/log((double)a)) #define Srand() srand((int)time(0)) #define random(number) (rand()%number) #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef vector<int> VI; typedef pair<int,int> PII; typedef vector<PII> VII; typedef vector<PII, int> VIII; typedef VI:: iterator IT; typedef map<string, int> Mps; typedef map<int, int> Mpi; typedef map<int, PII> Mpii; typedef map<PII, int> Mpiii; int n, n1, n2, n3, t1, t2, t3; const int maxn = 10000 + 100; int w[maxn]; int main() { scanf("%d%d%d%d%d%d%d", &n, &n1, &n2, &n3, &t1, &t2, &t3); Rep(i, 1, n+1) { int c = 0; if(i > n1) c = max(c, w[i-n1]+t1); if(i > n2) c = max(c, w[i-n2]+t2); if(i > n3) c = max(c, w[i-n3]+t3); w[i] = c; if(i == n) cout<<c+t1+t2+t3<<endl; } return 0; }
E题：给定3个串s1,s2,s3,对每一个l(1<=l<=min(|s1|,|s2|,|s3|))求三元组(i,j,k)的个数，三元组是指满足
##：s1[i,i+l-1], s2[j,j+l-1],s3[k,k+l-1]是完全相同字符串的条件。
分析：表示知道用后缀数组处理，然后就不知道怎么做了，看了题解还是有点不明白，先留个坑，
UPD:终于吧这道题弄懂了。
分析：题目是要统计对于长度为l的字串，三元组的(i, j, k)的个数，其中满足上面##的条件。很巧妙的用到了并查集的，我们可以利用后缀数组求出h数组。
将LCP值相同的对应的h下标，也就是h值相同的下标集中存储，然后按照h从大到小,依次将h数组中位于相邻位置，且LCP>=当前h值的下标用并查集合并在一起。
对于并查集中根节点rt, cnt[0][rt], cnt[1][rt], cnt[2][rt]表示LCP大于当前的h值中，所有的子串中出自第1,2,3个的有多少个，每次合并之前减去之前的值，利用合并之后的cnt数组进行更新。
代码：
1 #include <bits/stdc++.h> 2 #define in freopen("solve_in.txt", "r", stdin); 3 #define bug(x) printf("Line #%d: >>>>><<< ", (x)); 4 typedef long long LL; 5 6 using namespace std; 7 const LL M = 1000000007; 8 9 const int maxn = 3*(int)1e5+1000; 10 int s[maxn], sa[maxn], t[maxn], t2[maxn], c[maxn], h[maxn], rk[maxn]; 11 int n; 12 char str[3][maxn]; 13 LL cnt[3][maxn]; 14 int fa[maxn]; 15 int slen[3]; 16 int res[maxn]; 17 18 vector<int> g[maxn]; 19 20 void buildSa(int m) 21 { 22 int *x = t, *y = t2; 23 for(int i = 0; i < m; i++) c[i] = 0; 24 for(int i = 0; i < n; i++) c[x[i] = s[i]]++; 25 for(int i = 1; i < m; i++) c[i] += c[i-1]; 26 for(int i = n-1; i >= 0; i--) sa[--c[x[i]]] = i; 27 28 for(int k = 1; k <= n; k <<= 1) 29 { 30 int p = 0; 31 for(int i = n-k; i < n; i++) y[p++] = i; 32 33 for(int i = 0; i < m; i++) c[i] = 0; 34 for(int i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i]-k; 35 for(int i = 0; i < n; i++) c[x[y[i]]]++; 36 for(int i = 1; i < m; i++) c[i] += c[i-1]; 37 38 for(int i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i]; 39 swap(x, y); 40 p = 1; 41 x[sa[0]] = 0; 42 for(int i = 1; i < n; i++) 43 x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++; 44 if(p >= n) break; 45 m = p; 46 } 47 } 48 void getHeight() 49 { 50 int j, k = 0; 51 h[0] = 0; 52 for(int i = 0; i < n; i++) rk[sa[i]] = i; 53 for(int i = 0; i < n; i++) 54 { 55 56 if(k) k--; 57 if(rk[i] == 0) 58 continue; 59 j = sa[rk[i]-1]; 60 while(s[i+k] == s[j+k]) k++; 61 h[rk[i]] = k; 62 } 63 } 64 int findset(int x) 65 { 66 return x == fa[x] ? x : fa[x] = findset(fa[x]); 67 } 68 void Union(int x, int y) 69 { 70 int fx = findset(x); 71 int fy = findset(y); 72 if(fx == fy) 73 return; 74 for(int i = 0; i < 3; i++) 75 cnt[i][fx] += cnt[i][fy]; 76 fa[fy] = fx; 77 } 78 79 void init() 80 { 81 for(int i = 0; i < maxn; i++) 82 { 83 fa[i] = i; 84 } 85 int base = 0; 86 for(int k = 0; k < 3; k++) 87 { 88 for(int j = 0; j < slen[k]; j++) 89 cnt[k][j+base+k] = 1; 90 base += slen[k]; 91 } 92 } 93 94 void solve() 95 { 96 LL ans = 0; 97 for(int i = n-1; i > 0; i--) 98 { 99 int sz = g[i].size(); 100 for(int j = 0; j < sz; j++) 101 { 102 int v = g[i][j]; 103 int x = findset(sa[v-1]); 104 int y = findset(sa[v]); 105 ans -= cnt[0][x]*cnt[1][x]%M*cnt[2][x]%M; 106 if(ans < 0) 107 ans += M; 108 ans -= cnt[0][y]*cnt[1][y]%M*cnt[2][y]%M; 109 if(ans < 0) 110 ans += M; 111 Union(x, y); 112 x = findset(x); 113 ans = (ans+cnt[0][x]*cnt[1][x]%M*cnt[2][x]%M)%M; 114 115 } 116 117 res[i] = ans; 118 } 119 120 int len = min(slen[0], min(slen[1], slen[2])); 121 for(int i = 1; i <= len; i++) 122 printf("%d%c", res[i], i == len ? ' ' : ' '); 123 } 124 int main() 125 { 126 127 scanf("%s%s%s", str[0], str[1], str[2]); 128 for(int k = 0; k < 3; k++) 129 { 130 for(int i = 0; str[k][i]; i++) 131 { 132 s[n++] = str[k][i] - 'a' + 1; 133 slen[k]++; 134 } 135 s[n++] = 30+k; 136 } 137 s[n++] = 0; 138 init(); 139 buildSa(60); 140 getHeight(); 141 for(int i = 1; i < n; i++) 142 if(h[i]) 143 g[h[i]].push_back(i); 144 solve(); 145 return 0; 146 }
思路很类似的一道题：
F题：给定一个数组a[1...n]，为1....n的数的排列， n<=300000,是否存在两个数a，b，a+b/2在这数组中位于这两个数之间。实际就是判断是否三个数构成等差数列。
分析：这题要用线段树+hash来维护，首先容易知道，a,b,c构成等差那么 a，b之差与b,c之差是相等的，那么过程是这样的依次每次取出一个数a[i]，然后把a[i]和n-a[i]+1插入到两棵线段树中；如果不存在两个数与a[i]构成等差数列，例如数组中10个元素，对于2,5,8，5不能和2,8构成等差，2,8在数组中位置要么在5前面，要么在5后面，如果2,8在前面，那么将2,8以及10-2+1 = 9,10-8+1=3插入后，两棵线段树相应位置均会被增加相同元素，这样对应的线段树就应该都是序列2...8,那么比较两棵线段树相应区间hash值就可以了。同样对于2,8出现在5后面也可以判断。
一旦遇到hash值不同的情况说明存在等差数列。
代码：
1 //Template updates date: 20140718 2 #include <bits/stdc++.h> 3 #define esp 1e-6 4 #define inf 0x3f3f3f3f 5 #define pi acos(-1.0) 6 #define pb push_back 7 #define lson l, m, rt<<1 8 #define rson m+1, r, rt<<1|1 9 #define lowbit(x) (x&(-x)) 10 #define mp(a, b) make_pair((a), (b)) 11 #define bit(k) (1<<(k)) 12 #define iin freopen("pow.in", "r", stdin); 13 #define oout freopen("pow.out", "w", stdout); 14 #define in freopen("solve_in.txt", "r", stdin); 15 #define out freopen("solve_out.txt", "w", stdout); 16 #define bug puts("********))))))"); 17 #define Inout iin oout 18 #define inout in out 19 20 #define SET(a, v) memset(a, (v), sizeof(a)) 21 #define SORT(a) sort((a).begin(), (a).end()) 22 #define REV(a) reverse((a).begin(), (a).end()) 23 #define READ(a, n) {REP(i, n) cin>>(a)[i];} 24 #define REP(i, n) for(int i = 0; i < (n); i++) 25 #define VREP(i, n, base) for(int i = (n); i >= (base); i--) 26 #define Rep(i, base, n) for(int i = (base); i < (n); i++) 27 #define REPS(s, i) for(int i = 0; (s)[i]; i++) 28 #define pf(x) ((x)*(x)) 29 #define mod(n) ((n)) 30 #define Log(a, b) (log((double)b)/log((double)a)) 31 #define Srand() srand((int)time(0)) 32 #define random(number) (rand()%number) 33 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a) 34 35 using namespace std; 36 typedef long long LL; 37 typedef unsigned long long ULL; 38 typedef vector<int> VI; 39 typedef pair<int,int> PII; 40 typedef vector<PII> VII; 41 typedef vector<PII, int> VIII; 42 typedef VI:: iterator IT; 43 typedef map<string, int> Mps; 44 typedef map<int, int> Mpi; 45 typedef map<int, PII> Mpii; 46 typedef map<PII, int> Mpiii; 47 48 const int maxn = 300000 + 100; 49 const int L = 100003; 50 51 LL b[maxn]; 52 53 int n; 54 int a[maxn]; 55 LL ans; 56 struct SegTree { 57 LL h[maxn<<2]; 58 void build(int l, int r, int rt) { 59 if(l == r) { 60 h[rt] = 0; 61 return; 62 } 63 int m = (l+r)>>1; 64 build(lson); 65 build(rson); 66 } 67 void PushUp(int l, int r, int rt) { 68 int m = (l+r)>>1; 69 int ll = r-m; 70 h[rt] = h[rt<<1]*b[ll]+h[rt<<1|1]; 71 } 72 void update(int l, int r, int rt, int pos) { 73 if(pos == l && pos == r) { 74 h[rt] = b[1]; 75 return; 76 } 77 int m = (l+r)>>1; 78 if(pos <= m) 79 update(lson, pos); 80 else update(rson, pos); 81 PushUp(l, r, rt); 82 } 83 void query(int L, int R, int l, int r, int rt) { 84 if(L > R) 85 return; 86 if(L <= l && R >= r) { 87 ans = ans*b[r-l+1]+h[rt]; 88 return; 89 } 90 int m = (l+r)>>1; 91 if(L <= m) 92 query(L, R, lson); 93 if(R > m) 94 query(L, R, rson); 95 } 96 } t1, t2; 97 int main() { 98 99 scanf("%d", &n); 100 b[0] = 1; 101 Rep(i, 1, n+1) { 102 scanf("%d", a+i); 103 b[i] = b[i-1]*L; 104 } 105 t1.build(1, n, 1); 106 t2.build(1, n, 1); 107 Rep(i, 1, n+1) { 108 int v = a[i]; 109 int l = min(v-1, n-v); 110 ans = 0; 111 t1.query(v-l, v-1, 1, n, 1); 112 LL tmp = ans; 113 ans = 0; 114 t2.query(n-v-l+1, n-v, 1, n, 1); 115 if(tmp != ans) { 116 puts("YES"); 117 return 0; 118 } 119 t1.update(1, n, 1, v); 120 t2.update(1, n, 1, n-v+1); 121 } 122 puts("NO"); 123 return 0; 124 }