标签：span eve 区间修改 16px lap cout ffffff col query
思路：二分答案水过
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 100100 using namespace std; int n,m,tx,ty,ans; int cntx[MAXN],cnty[MAXN]; struct nond{ int x,y,d; double k; }cnt[MAXN]; bool check(int mid){ double dy=cnt[mid].k*tx+cnt[mid].d; if(dy<=ty) return true; else return false; } int main(){ freopen("geometry.in","r",stdin); freopen("geometry.out","w",stdout); scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&cntx[i]); for(int i=1;i<=n;i++) scanf("%d",&cnty[i]); sort(cntx+1,cntx+1+n); sort(cnty+1,cnty+1+n); for(int i=1;i<=n;i++){ cnt[i].x=cntx[i]; cnt[i].y=cnty[i]; cnt[i].k=-1*(cnty[i]*1.0/cntx[i]*1.0); cnt[i].d=cnty[i]; } scanf("%d",&m); for(int i=1;i<=m;i++){ scanf("%d%d",&tx,&ty); int l=0,r=n,mid; while(l<=r){ mid=(l+r)/2; if(check(mid)) l=mid+1,ans=mid; else r=mid-1; } cout<<ans<<endl; } }
思路：贪心，水了45分，据说大佬的贪心可以水85分的。
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 50 using namespace std; int t,sum,ans; int need[MAXN],people[MAXN],vis[MAXN]; int main(){ freopen("cashier.in","r",stdin); freopen("cashier.out","w",stdout); scanf("%d",&t); while(t--){ ans=0; bool flag=0; memset(vis,0,sizeof(vis)); memset(need,0,sizeof(need)); memset(people,0,sizeof(people)); for(int i=1;i<=24;i++){ scanf("%d",&need[i]); need[24+i]=need[i]; } for(int i=1;i<=24;i++){ scanf("%d",&people[i]); people[24+i]=people[i]; } for(int i=25;i<=48;i++){ if(need[i]!=0){ sum=0; for(int j=i-7;j<=i;j++) if(vis[j]!=0) sum+=vis[j]; if(sum>=need[i]) continue; if(i-7>=25){ for(int j=i;j>=i-7;j--) if(vis[j]!=people[j]&&people[j]!=0){ if(sum==need[i]) break; if(sum+people[j]-vis[j]<=need[i]) sum+=people[j]-vis[j],vis[j]=people[j]; else if(sum+people[j]-vis[j]>need[i]) vis[j]=need[i]-sum,sum=need[i]; } } else{ for(int j=i-7;j<=i;j++) if(vis[j]!=people[j]&&people[j]!=0){ if(sum==need[i]) break; if(sum+people[j]-vis[j]<=need[i]){ sum+=people[j]-vis[j],vis[j]=people[j]; if(j<=24) sum+=people[j]-vis[j],vis[j+24]=people[j]; } else if(sum+people[j]-vis[j]>need[i]){ vis[j]=need[i]-sum,sum=need[i]; if(j<=24) vis[j+24]=need[i]-sum,sum=need[i]; } } } if(sum<need[i]){ cout<<"-1";flag=1;break; } } } for(int i=25;i<=48;i++) ans+=vis[i]; if(!flag) cout<<ans<<endl; } } /* 1 0 0 0 0 0 0 0 0 4 5 2 2 2 6 3 2 4 3 2 1 1 2 1 0 1 1 1 0 0 0 1 1 3 1 1 0 0 2 0 0 0 3 1 1 1 0 0 2 1 0 1 0 0 0 0 0 0 2 2 1 5 2 0 3 1 1 4 1 4 2 6 3 3 1 1 1 0 1 0 0 0 0 1 3 2 0 0 4 0 3 0 1 0 0 2 0 0 */
正解思路：查分约束
记录前缀和sum[i](前缀和是指最终应该得到的答案的前缀和)
当i>=8时 应该满足条件：s[i]-s[i-8]>=a[i];
0<=s[i]-s[i-1]<=b[i];
当i< 8时 应该满足条件：s[23]-s[16+i]+s[i]>=a[i];（下标从0开始）
但是第3个式子并不满足差分约束的性质。
我们可以发现第三个式子中的s[23]具有单调性，所以可以从小到大枚举s[23]是几，第一个符合条件的一定是最小的（s[23](也就是最终的答案是几),二分s[23]的值也可以），然后就可以转为符合差分约束的式子：
s[i]-s[16+i]>=a[i]-s[23];
然后spfa判负环，跑差分约束即可。
#include<queue> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 50 using namespace std; queue<int>que; int t,tot,ans; int a[MAXN],b[MAXN]; int dis[MAXN],vis[MAXN]; int to[MAXN*2],cap[MAXN*2],net[MAXN*2],head[MAXN*2]; void add(int u,int v,int w){ to[++tot]=v;cap[tot]=w;net[tot]=head[u];head[u]=tot; } bool spfa(int s){ memset(vis,0,sizeof(vis)); memset(dis,-0x7f7f7f7f,sizeof(dis)); while(!que.empty()) que.pop(); que.push(s); dis[s]=0;vis[s]=1; while(!que.empty()){ int now=que.front(); que.pop(); vis[now]=0; if(dis[now]>1000) return false; for(int i=head[now];i;i=net[i]) if(dis[to[i]]<dis[now]+cap[i]){ dis[to[i]]=dis[now]+cap[i]; if(!vis[to[i]]){ vis[to[i]]=1; que.push(to[i]); } } } return true; } bool judge(int ans){ memset(head,0,sizeof(head));tot=0; for(int i=9;i<=24;i++) add(i-8,i,a[i]); for(int i=1;i<=8;i++) add(i+16,i,a[i]-ans); for(int i=1;i<=24;i++) add(i-1,i,0); for(int i=1;i<=24;i++) add(i,i-1,-b[i]); add(0,24,ans); add(24,0,-ans); return spfa(0); } int main(){ freopen("cashier.in","r",stdin); freopen("cashier.out","w",stdout); scanf("%d",&t); while(t--){ for(int i=1;i<=24;i++) scanf("%d",&a[i]); for(int i=1;i<=24;i++) scanf("%d",&b[i]); ans=0; while(1){ if(++ans>1000){ ans=-1; break; } if(judge(ans)) break; } cout<<ans<<endl; } } /* 1 0 0 0 0 0 0 0 0 2 2 1 5 2 0 3 1 1 4 1 4 2 6 3 3 1 1 1 0 1 0 0 0 0 1 3 2 0 0 4 0 3 0 1 0 0 2 0 0 */
思路：
自己写的80分暴力的线段树，然并卵，爆零了，QwQ求大佬找茬...
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 50100 #define mod 1000000007 using namespace std; int n,m; struct nond{ int l,r; long long sum; long long flag1,flag2; }tree[MAXN*4]; void up(int now){ tree[now].sum=(tree[now*2].sum+tree[now*2+1].sum)%mod; } void build(int now,int l,int r){ tree[now].l=l; tree[now].r=r; tree[now].flag2=1; if(tree[now].l==tree[now].r){ scanf("%d",&tree[now].sum); return ; } int mid=(tree[now].l+tree[now].r)/2; build(now*2,l,mid); build(now*2+1,mid+1,r); up(now); } void down(int now){ tree[now*2].flag2=(tree[now].flag2*tree[now*2].flag2)%mod; tree[now*2+1].flag2=(tree[now].flag2*tree[now*2+1].flag2)%mod; tree[now*2].flag1=(tree[now].flag2*tree[now*2].flag1%mod+tree[now].flag1)%mod; tree[now*2+1].flag1=(tree[now].flag2*tree[now*2+1].flag1%mod+tree[now].flag1)%mod; tree[now*2].sum=(tree[now].flag2*tree[now*2].sum%mod+tree[now].flag1*(tree[now*2].r-tree[now*2].l+1)%mod)%mod; tree[now*2+1].sum=(tree[now].flag2*tree[now*2+1].sum%mod+tree[now].flag1*(tree[now*2+1].r-tree[now*2+1].l+1)%mod)%mod; tree[now].flag2=1; tree[now].flag1=0; } void change1(int now,int l,int r,int k){ if(tree[now].l==l&&tree[now].r==r){ tree[now].flag1+=k; tree[now].sum+=1ll*(tree[now].r-tree[now].l+1)*k; return ; } down(now); int mid=(tree[now].l+tree[now].r)/2; if(r<=mid) change1(now*2,l,r,k); else if(l>mid) change1(now*2+1,l,r,k); else{ change1(now*2,l,mid,k); change1(now*2+1,mid+1,r,k); } up(now); } void change2(int now,int l,int r,int v){ if(tree[now].l==l&&tree[now].r==r){ tree[now].flag2=(tree[now].flag2*v)%mod; tree[now].flag1=(tree[now].flag1*v)%mod; tree[now].sum=(tree[now].sum*v)%mod; return ; } down(now); int mid=(tree[now].l+tree[now].r)/2; if(r<=mid) change2(now*2,l,r,v); else if(l>mid) change2(now*2+1,l,r,v); else{ change2(now*2,l,mid,v); change2(now*2+1,mid+1,r,v); } up(now); } long long query(int now,int l,int r){ if(tree[now].l==l&&tree[now].r==r) return tree[now].sum%mod; down(now); int mid=(tree[now].l+tree[now].r)/2; if(r<=mid) return query(now*2,l,r); else if(l>mid) return query(now*2+1,l,r); else return query(now*2,l,mid)+query(now*2+1,mid+1,r); } int main(){ freopen("game.in","r",stdin); freopen("game.out","w",stdout); scanf("%d%d",&n,&m); build(1,1,n); for(int i=1;i<=m;i++){ int a,b,c,d; scanf("%d%d%d",&a,&b,&c); if(a==1){ scanf("%d",&d); change1(1,b,c,d); } if(a==2) change2(1,b,c,-1); if(a==3){ scanf("%d",&d); if(d==1){ long long ans=0; ans=query(1,b,c); if(ans<0) ans+=mod; cout<<ans<<endl; } if(d==2){ long long ans=0; for(int i=b;i<c;i++) ans=(ans+query(1,i+1,c)*query(1,i,i))%mod; if(ans<0) ans+=mod; cout<<ans<<endl; } } } } /* 10 9 3 6 7 4 6 1 6 7 2 6 3 5 7 3 1 1 7 -9 1 2 3 5 3 2 6 1 2 5 8 3 5 7 3 2 2 3 3 1 10 2 3 1 2 2 */
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=1时线段树区间求和区间修改，我只会这个....
线段树区间合并操作。
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)
只懂思路，代码....果断粘std
#include <cstdio> #include <cstdlib> #define MOD 1000000007 #define N 100005 typedef long long LL; using namespace std; struct Node { LL f[11]; } node[N * 4]; LL a[N], lazy1[N * 4]; bool lazy2[N * 4]; LL C[N][11]; Node merge(Node lc, Node rc) { Node o; o.f[0] = 1; for (int i = 1; i <= 10; i++) { o.f[i] = 0; for (int j = 0; j <= i; j++) o.f[i] = (o.f[i] + lc.f[j] * rc.f[i - j] % MOD) % MOD; } return o; } void build(int o, int l, int r) { if (l == r) { for (int i = 0; i <= 10; i++) node[o].f[i] = 0; node[o].f[0] = 1; node[o].f[1] = (a[l] % MOD + MOD) % MOD; return ; } int mid = (l + r) >> 1; build(o * 2, l, mid); build(o * 2 + 1, mid + 1, r); node[o] = merge(node[o * 2], node[o * 2 + 1]); return ; } void update1(int o, int l, int r, int c) { int len = r - l + 1; LL ff[11]; for (int i = 0; i <= 10; i++) ff[i] = node[o].f[i]; for (int i = 1; i <= 10; i++) { node[o].f[i] = 0; LL t = 1; for (int j = 0; j <= i; j++) { LL tmp = ff[i - j] * C[len - (i - j)][j] % MOD * t % MOD; node[o].f[i] = (node[o].f[i] + tmp) % MOD; t = t * c % MOD; } } return ; } void push_down(int o, int l, int r) { int mid = (l + r) >> 1; if (lazy1[o]) { if (lazy2[o * 2]) lazy1[o * 2] = (lazy1[o * 2] + MOD - lazy1[o]) % MOD; else lazy1[o * 2] = (lazy1[o * 2] + lazy1[o]) % MOD; if (lazy2[o * 2 + 1]) lazy1[o * 2 + 1] = (lazy1[o * 2 + 1] + MOD - lazy1[o]) % MOD; else lazy1[o * 2 + 1] = (lazy1[o * 2 + 1] + lazy1[o]) % MOD; update1(o * 2, l, mid, lazy1[o]); update1(o * 2 + 1, mid + 1, r, lazy1[o]); lazy1[o] = 0; } if (lazy2[o]) { lazy2[o * 2] ^= 1; lazy2[o * 2 + 1] ^= 1; for (int j = 1; j <= 10; j += 2) { node[o * 2].f[j] = MOD - node[o * 2].f[j]; node[o * 2 + 1].f[j] = MOD - node[o * 2 + 1].f[j]; } lazy2[o] = 0; } } void modify1(int o, int l, int r, int ll, int rr, int c) { if (ll <= l && rr >= r) { if (lazy2[o]) lazy1[o] = (lazy1[o] + MOD - c) % MOD; else lazy1[o] = (lazy1[o] + c) % MOD; update1(o, l, r, c); return ; } int mid = (l + r) >> 1; push_down(o, l, r); if (ll <= mid) modify1(o * 2, l, mid, ll, rr, c); if (rr > mid) modify1(o * 2 + 1, mid + 1, r, ll, rr, c); node[o] = merge(node[o * 2], node[o * 2 + 1]); return ; } void modify2(int o, int l, int r, int ll, int rr) { if (ll <= l && rr >= r) { for (int i = 1; i <= 10; i += 2) node[o].f[i] = MOD - node[o].f[i]; lazy2[o] ^= 1; return ; } int mid = (l + r) >> 1; push_down(o, l, r); if (ll <= mid) modify2(o * 2, l, mid, ll, rr); if (rr > mid) modify2(o * 2 + 1, mid + 1, r, ll, rr); node[o] = merge(node[o * 2], node[o * 2 + 1]); return ; } Node query(int o, int l, int r, int ll, int rr) { if (ll <= l && rr >= r) return node[o]; int mid = (l + r) >> 1; push_down(o, l, r); if (rr <= mid) return query(o * 2, l, mid, ll, rr); if (ll > mid) return query(o * 2 + 1, mid + 1, r, ll, rr); Node lc = query(o * 2, l, mid, ll, rr); Node rc = query(o * 2 + 1, mid + 1, r, ll, rr); return merge(lc, rc); } int main(int argc, char ** argv) { freopen("game.in", "r", stdin); freopen("game.out", "w", stdout); int n, m; scanf("%d %d", &n, &m); C[0][0] = 1; for (int i = 1; i <= n; i++) { C[i][0] = 1; for (int j = 1; j <= 10; j++) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD; } for (int i = 1; i <= n; i++) scanf("%d", &a[i]); build(1, 1, n); for (int i = 1; i <= m; i++) { int l, r, opt; scanf("%d%d%d",&opt, &l, &r); if (opt == 1) { int c; scanf("%d", &c); c = (c % MOD + MOD) % MOD; modify1(1, 1, n, l, r, c); } else if (opt == 2) { modify2(1, 1, n, l, r); } else { int k; scanf("%d", &k); Node o = query(1, 1, n, l, r); printf("%d\n", o.f[k] % MOD); } } return 0; }
标签：span eve 区间修改 16px lap cout ffffff col query
原文地址：http://www.cnblogs.com/cangT-Tlan/p/7677709.html
