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先排序 因为题目保证一定存在一种情况使得线段两两不相交 如下图
我们观察可以发现如果一个点与(0,0)，连线 如果线段与第i条线段相交 那么一定会与第i-1条线段相交
那么就可以二分判断这条线段与多少条已知线段相交
1 #include <cstdio> 2 #include <cctype> 3 #include <algorithm> 4 5 const int MAXN=100010; 6 7 int n,m,xx,yy; 8 9 int x[MAXN],y[MAXN]; 10 11 inline void read(int&x) { 12 int f=1;register char c=getchar(); 13 for(x=0;!isdigit(c);c=='-'&&(f=-1),c=getchar()); 14 for(;isdigit(c);x=x*10+c-48,c=getchar()); 15 x=x*f; 16 } 17 18 bool check(int p) { 19 double k=(double)y[p]/x[p]; 20 double _y=(double)-k*xx+y[p]; 21 if(yy<_y) return true; 22 else return false; 23 } 24 25 int hh() { 26 freopen("geometry.in","r",stdin); 27 freopen("geometry.out","w",stdout); 28 29 read(n); 30 for(int i=1;i<=n;++i) read(x[i]); 31 for(int i=1;i<=n;++i) read(y[i]); 32 std::sort(x+1,x+1+n); 33 std::sort(y+1,y+1+n); 34 read(m); 35 for(int i=1;i<=m;++i) { 36 read(xx);read(yy); 37 int l=0,r=n+1,ans=0; 38 while(l+1<r) { 39 int mid=(l+r)>>1; 40 if(check(mid)) r=mid; 41 else l=mid; 42 } 43 ans+=l; 44 printf("%d ",ans); 45 } 46 return 0; 47 } 48 49 int sb=hh(); 50 int main(int argc,char**argv) {;}
差分约束
记录前缀和S[i],i>=8时首先要满足条件 s[i]-s[i-8]>=a[i]
0<=s[i]-s[i-1]<=b[i]
当i<8时有s[23]-s[16+i]+s[i]>=a[i]
因为第三个式子不满足差分约束形式，可以看出s[23]有单调性，可以二分…所以可以二分答案当常量处理
Spfa负环则无解
还有一种网络流做法，表示很懵逼。
1 #include <cstdio> 2 #include <cstdlib> 3 #include <queue> 4 using namespace std; 5 6 #define N 33 7 8 struct edge{ 9 int t, n, l; 10 }e[N * N]; 11 12 int h[N]; 13 int tote; 14 15 void adde(int u, int v, int l) { 16 e[++tote].t = v; 17 e[tote].l = l; 18 e[tote].n = h[u]; 19 h[u] = tote; 20 return ; 21 } 22 23 int d[N]; 24 bool inq[N]; 25 26 bool spfa() { 27 queue<int> Q; 28 for (int i = 0; i <= 24; i++) d[i] = -1e9, inq[i] = false; 29 d[0] = 0; inq[0] = true; Q.push(0); 30 while (!Q.empty()) { 31 int u = Q.front(); Q.pop(); inq[u] = false; 32 if (d[u] > 1000) return false; 33 for (int i = h[u]; i; i = e[i].n) { 34 int v = e[i].t, l = e[i].l; 35 if (d[v] < d[u] + l) { 36 d[v] = d[u] + l; 37 if (!inq[v]) { 38 inq[v] = true; 39 Q.push(v); 40 } 41 } 42 } 43 } 44 return true; 45 } 46 47 int a[30], b[30]; 48 49 bool solve(int ans) { 50 for (int i = 0; i <= 24; i++) h[i] = 0; 51 tote = 0; 52 for (int i = 9; i <= 24; i++) adde(i - 8, i, a[i]); 53 for (int i = 1; i <= 8; i++) adde(i + 16, i, a[i] - ans); 54 for (int i = 1; i <= 24; i++) adde(i - 1, i, 0); 55 for (int i = 1; i <= 24; i++) adde(i, i - 1, -b[i]); 56 adde(0, 24, ans); adde(24, 0, -ans); 57 return spfa(); 58 } 59 60 int main(int argc, char ** argv) { 61 freopen("cashier.in", "r", stdin); 62 freopen("cashier.out", "w", stdout); 63 int test; 64 scanf("%d", &test); 65 for (int i = 1; i <= test; i++) { 66 for (int i = 1; i <= 24; i++) scanf("%d", &a[i]); 67 for (int i = 1; i <= 24; i++) scanf("%d", &b[i]); 68 int ans = 0; 69 while (true) { 70 if (++ans > 1000) { 71 ans = -1; break; 72 } 73 if (solve(ans)) break; 74 } 75 printf("%d ", ans); 76 } 77 fclose(stdin); 78 fclose(stdout); 79 return 0; 80 }
线段树
部分分可以dp f[i][j]表示前i个数选了j个的答案
f[i][j]=f[i-1][j]+f[i-1][j-1]*a[i] (i选不选)
k比较小，所以线段树每个节点维护一个区间答案记为f[i]
考虑一段区间i,左边取j个右边就取i-j个 答案是每个方案的左边乘右边的和。
就是i左儿子f[j]和右边的f[i-j] 所以f[i]=Σ(j=0~i) lc f[j]*rc f[i-j]
考虑取反操作，i是奇数就取反，偶数无影响(因为是相乘)
考虑区间加， 开始f[i] 是 a1*a2……an 后来是(a1+c)*(a2+c)……(an+c)
考虑类似二项式定理,当上述a1~an n个方案如果取了j个了，分别为al1,al2……alj
那考虑最后答案，如果已经选了j个方案是(al1+c)(al2+c)……(alj+c)再还能选i-j个 最后答案是C(len-i,i-j)*f[j]*c^(i-j)
复杂度 O(k^2*nlogn)
1 #include <cstdio> 2 #include <cstdlib> 3 #define MOD 1000000007 4 #define N 100005 5 typedef long long LL; 6 using namespace std; 7 struct Node { 8 LL f[11]; 9 }node[N * 4]; 10 LL a[N], lazy1[N * 4]; 11 bool lazy2[N * 4]; 12 LL C[N][11]; 13 14 Node merge(Node lc, Node rc) { 15 Node o; 16 o.f[0] = 1; 17 for (int i = 1; i <= 10; i++) { 18 o.f[i] = 0; 19 for (int j = 0; j <= i; j++) 20 o.f[i] = (o.f[i] + lc.f[j] * rc.f[i - j] % MOD) % MOD; 21 } 22 return o; 23 } 24 25 void build(int o, int l, int r) { 26 if (l == r) { 27 for (int i = 0; i <= 10; i++) node[o].f[i] = 0; 28 node[o].f[0] = 1; 29 node[o].f[1] = (a[l] % MOD + MOD) % MOD; 30 return ; 31 } 32 int mid = (l + r) >> 1; 33 build(o * 2, l, mid); 34 build(o * 2 + 1, mid + 1, r); 35 node[o] = merge(node[o * 2], node[o * 2 + 1]); 36 return ; 37 } 38 39 void update1(int o, int l, int r, int c) { 40 int len = r - l + 1; 41 LL ff[11]; 42 for (int i = 0; i <= 10; i++) ff[i] = node[o].f[i]; 43 for (int i = 1; i <= 10; i++) { 44 node[o].f[i] = 0; 45 LL t = 1; 46 for (int j = 0; j <= i; j++) { 47 LL tmp = ff[i - j] * C[len - (i - j)][j] % MOD * t % MOD; 48 node[o].f[i] = (node[o].f[i] + tmp) % MOD; 49 t = t * c % MOD; 50 } 51 } 52 return ; 53 } 54 55 void push_down(int o, int l, int r) { 56 int mid = (l + r) >> 1; 57 if (lazy1[o]) { 58 if (lazy2[o * 2]) 59 lazy1[o * 2] = (lazy1[o * 2] + MOD - lazy1[o]) % MOD; 60 else 61 lazy1[o * 2] = (lazy1[o * 2] + lazy1[o]) % MOD; 62 if (lazy2[o * 2 + 1]) 63 lazy1[o * 2 + 1] = (lazy1[o * 2 + 1] + MOD - lazy1[o]) % MOD; 64 else 65 lazy1[o * 2 + 1] = (lazy1[o * 2 + 1] + lazy1[o]) % MOD; 66 update1(o * 2, l, mid, lazy1[o]); 67 update1(o * 2 + 1, mid + 1, r, lazy1[o]); 68 lazy1[o] = 0; 69 } 70 if (lazy2[o]) { 71 lazy2[o * 2] ^= 1; 72 lazy2[o * 2 + 1] ^= 1; 73 for (int j = 1; j <= 10; j += 2) { 74 node[o * 2].f[j] = MOD - node[o * 2].f[j]; 75 node[o * 2 + 1].f[j] = MOD - node[o * 2 + 1].f[j]; 76 } 77 lazy2[o] = 0; 78 } 79 } 80 81 void modify1(int o, int l, int r, int ll, int rr, int c) { 82 if (ll <= l && rr >= r) { 83 if (lazy2[o]) lazy1[o] = (lazy1[o] + MOD - c) % MOD; 84 else lazy1[o] = (lazy1[o] + c) % MOD; 85 update1(o, l, r, c); 86 return ; 87 } 88 int mid = (l + r) >> 1; 89 push_down(o, l, r); 90 if (ll <= mid) modify1(o * 2, l, mid, ll, rr, c); 91 if (rr > mid) modify1(o * 2 + 1, mid + 1, r, ll, rr, c); 92 node[o] = merge(node[o * 2], node[o * 2 + 1]); 93 return ; 94 } 95 96 void modify2(int o, int l, int r, int ll, int rr) { 97 if (ll <= l && rr >= r) { 98 for (int i = 1; i <= 10; i += 2) node[o].f[i] = MOD - node[o].f[i]; 99 lazy2[o] ^= 1; 100 return ; 101 } 102 int mid = (l + r) >> 1; 103 push_down(o, l, r); 104 if (ll <= mid) modify2(o * 2, l, mid, ll, rr); 105 if (rr > mid) modify2(o * 2 + 1, mid + 1, r, ll, rr); 106 node[o] = merge(node[o * 2], node[o * 2 + 1]); 107 return ; 108 } 109 110 Node query(int o, int l, int r, int ll, int rr) { 111 if (ll <= l && rr >= r) 112 return node[o]; 113 int mid = (l + r) >> 1; 114 push_down(o, l, r); 115 if (rr <= mid) return query(o * 2, l, mid, ll, rr); 116 if (ll > mid) return query(o * 2 + 1, mid + 1, r, ll, rr); 117 Node lc = query(o * 2, l, mid, ll, rr); 118 Node rc = query(o * 2 + 1, mid + 1, r, ll, rr); 119 return merge(lc, rc); 120 } 121 122 int main(int argc, char ** argv) { 123 freopen("game.in", "r", stdin); 124 freopen("game.out", "w", stdout); 125 int n, m; 126 scanf("%d %d", &n, &m); 127 C[0][0] = 1; 128 for (int i = 1; i <= n; i++) { 129 C[i][0] = 1; 130 for (int j = 1; j <= 10; j++) 131 C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD; 132 } 133 for (int i = 1; i <= n; i++) 134 scanf("%d", &a[i]); 135 build(1, 1, n); 136 for (int i = 1; i <= m; i++) { 137 138 int l, r, opt; 139 scanf("%d%d%d",&opt, &l, &r); 140 if (opt == 1) { 141 int c; 142 scanf("%d", &c); 143 c = (c % MOD + MOD) % MOD; 144 modify1(1, 1, n, l, r, c); 145 } 146 else if (opt == 2) { 147 modify2(1, 1, n, l, r); 148 } 149 else { 150 int k; 151 scanf("%d", &k); 152 Node o = query(1, 1, n, l, r); 153 printf("%d ", o.f[k] % MOD); 154 } 155 } 156 return 0; 157 }