1002 小洛洛
开场挂了N发插入排序的我（ 最后还是要靠小洛洛写暴力
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <cmath>
5 #include <cstdlib>
6 #include <algorithm>
7 using namespace std;
8
9 const int N = 12;
10
11 int min_(char s[], int mul, int k, bool fst = false) {
12 if(!*s)
13 return 0;
14 char x = *s;
15 if(k)
16 for(char *p = s; *p; ++p)
17 if(!fst || *p != ‘0‘)
18 x = min(x, *p);
19 int mn = 1000000000;
20 if(*s == x)
21 mn = min_(s + 1, mul / 10, k);
22 else
23 for(char *p = s + 1; *p; ++p)
24 if(*p == x) {
25 swap(*s, *p);
26 mn = min(mn, min_(s + 1, mul / 10, k - 1));
27 swap(*s, *p);
28 }
29 return (x - ‘0‘) * mul + mn;
30 }
31 int max_(char s[], int mul, int k) {
32 if(!*s)
33 return 0;
34 char x = *s;
35 if(k)
36 for(char *p = s; *p; ++p)
37 x = max(x, *p);
38 int mx = 0;
39 if(*s == x)
40 mx = max_(s + 1, mul / 10, k);
41 else
42 for(char *p = s + 1; *p; ++p)
43 if(*p == x) {
44 swap(*s, *p);
45 mx = max(mx, max_(s + 1, mul / 10, k - 1));
46 swap(*s, *p);
47 }
48 return (x - ‘0‘) * mul + mx;
49 }
50
51 int main() {
52 int t;
53 scanf("%d", &t);
54 while(t--) {
55 int k;
56 char s[N];
57 scanf("%s%d", s, &k);
58 int mul = 1;
59 for(char *p = s + 1; *p; ++p)
60 mul *= 10;
61 int mn = min_(s, mul, k, true);
62 int mx = max_(s, mul, k);
63 printf("%d %d\n", mn, mx);
64 }
65 return 0;
66 }
1005 BPM136
用圆的面积交模板改一下就好了
1 /* ***********************************************
2 Author :BPM136
3 Created Time :2018/8/6 12:38:56
4 File Name :1005.cpp
5 ************************************************ */
6
7 #include<iostream>
8 #include<cstdio>
9 #include<algorithm>
10 #include<cstdlib>
11 #include<cmath>
12 #include<cstring>
13 #include<vector>
14 using namespace std;
15
16 const int N = 105;
17 const double pi = acos(-1.0);
18
19 struct point {
20 double x,y;
21 };
22
23 struct circle {
24 int x,y;
25 int R;
26 }c[N];
27 int m,R;
28
29 double cirinter(int r1,int x2,int y2,int r2) {
30 double d,s,t,a1,a2,s1,s2,s3;
31 d=sqrt((-x2)*(-x2)+(-y2)*(-y2));
32 if(d>=r1+r2)return 0;
33 else if(d<r1-r2) return 0;
34 else
35 {
36 a1=acos((r1*r1+d*d-r2*r2)/(2*r1*d));
37 a2=acos((r2*r2+d*d-r1*r1)/(2*r2*d));
38 s1=2*a1*r1;
39 s2=2*a2*r2;
40 s=s2-s1;
41 }
42 return s;
43 }
44
45 int main() {
46 int T;
47 scanf("%d",&T);
48 while(T--) {
49 scanf("%d%d",&m,&R);
50 for(int i=1;i<=m;i++) scanf("%d%d%d",&c[i].x,&c[i].y,&c[i].R);
51 double ans=2*pi*R;
52 for(int i=1;i<=m;i++) ans+=cirinter(R,c[i].x,c[i].y,c[i].R);
53 printf("%.10f\n",ans);
54 }
55 return 0;
56 }
1007 BPM136
正解st表下压待补
线段树据说势能分析一下是对的？
线段树版本
1 /* ***********************************************
2 Author :BPM136
3 Created Time :2018/8/6 17:45:43
4 File Name :1007G.cpp
5 ************************************************ */
6
7 #include<iostream>
8 #include<cstdio>
9 #include<algorithm>
10 #include<cstdlib>
11 #include<cmath>
12 #include<cstring>
13 #include<vector>
14 using namespace std;
15
16 typedef unsigned int UI;
17 typedef long long ll;
18
19 UI X,Y,Z;
20
21 UI RNG61() {
22 X=X^(X<<11);
23 X=X^(X>>4);
24 X=X^(X<<5);
25 X=X^(X>>14);
26 UI W=X^(Y^Z);
27 X=Y;
28 Y=Z;
29 Z=W;
30 return Z;
31 }
32
33 const int N = 100005;
34 const unsigned int MOD = (1U<<30);
35
36 int n,m;
37
38 #define LSON (k<<1)
39 #define RSON (k<<1|1)
40 #define MID ((l+r)>>1)
41
42 int mx[N<<2];
43
44 void update(int k,int l,int r,int ll,int rr,int val) {
45 if(mx[k]>=val) return ;
46 if(l==ll && r==rr) {
47 mx[k]=val;
48 return ;
49 }
50 int mid=MID;
51 if(rr<=mid) update(LSON,l,mid,ll,rr,val);
52 else if(ll>mid) update(RSON,mid+1,r,ll,rr,val);
53 else {
54 update(LSON,l,mid,ll,mid,val);
55 update(RSON,mid+1,r,mid+1,rr,val);
56 }
57 }
58
59 int query(int k,int pos) {
60 int l=1,r=n;
61 int ret=0;
62 while(l!=r) {
63 ret=max(ret,mx[k]);
64 int mid=MID;
65 if(pos<=mid) {
66 r=mid;
67 k=LSON;
68 } else {
69 l=mid+1;
70 k=RSON;
71 }
72 }
73 ret=max(ret,mx[k]);
74 return ret;
75 }
76
77 int main() {
78 int T;
79 cin>>T;
80 while(T--) {
81 cin>>n>>m>>X>>Y>>Z;
82 memset(mx,0,sizeof(mx));
83 for(int i=1;i<=m;i++) {
84 UI x=RNG61();
85 UI y=RNG61();
86 UI w=RNG61();
87 UI l,r,v;
88 l=min(x%n+1, y%n+1);
89 r=max(x%n+1, y%n+1);
90 while(w>MOD) w-=MOD;
91 update(1,1,n,l,r,w);
92 }
93 ll ans=0;
94 //for(int i=1;i<=n;i++) cerr<<query(1,i)<<‘ ‘; cerr<<endl;
95 for(int i=1;i<=n;i++) ans=ans^((ll)query(1,i)*i);
96 cout<<ans<<endl;
97 }
98 return 0;
99 }
1008 小洛洛
dp 枚举翻转的值域dp就好了
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <cmath>
5 #include <cstdlib>
6 #include <algorithm>
7 using namespace std;
8
9 template<int BI, int BO>
10 struct IO_ {
11 char bufi[BI + 1], *qi, bufo[BO], *qo;
12 IO_(): qi(bufi + BI), qo(bufo) {}
13 ~IO_() { o_(EOF); }
14 int i_() {
15 if(qi == bufi + BI) bufi[fread(bufi, 1, BI, stdin)] = ‘\0‘, qi = bufi;
16 return *qi ? *qi++ : EOF;
17 }
18 void o_(int c) {
19 if(c != EOF) *qo++ = c;
20 if(qo == bufo + BO || (c == EOF && qo != bufo))
21 fwrite(bufo, qo - bufo, 1, stdout), qo = bufo;
22 }
23 IO_ &operator>>(int &ret) {
24 ret = 0; int c;
25 while(isspace(c = i_()));
26 bool bm = c == ‘-‘ ? (c = i_(), true) : false;
27 while(ret = ret * 10 - (c - ‘0‘), isdigit(c = i_()));
28 if(!bm) ret = -ret;
29 return *this;
30 }
31 IO_ &operator>>(char *p) {
32 int c; while(isspace(c = i_()));
33 do *p++ = c; while(!isspace(c = i_()));
34 *p = ‘\0‘;
35 return *this;
36 }
37 IO_ &operator<<(char c) { return o_(c), *this; }
38 IO_ &operator<<(char *p) { while(*p) o_(*p++); return *this; }
39 IO_ &operator<<(int x) {
40 char s[12] = {}, *p = s + 11;
41 bool bm = x < 0 ? true : (x = -x, false);
42 while(*--p = ‘0‘ - x % 10, x /= 10);
43 if(bm) *--p = ‘-‘;
44 while(*p) o_(*p++);
45 return *this;
46 }
47 };
48
49 const int N = 100002;
50 char s[N];
51 int l[N][10], r[N][10];
52 int f[N][10][2];
53
54 void init_(int n) {
55 for(int i = 0; i < n; ++i)
56 for(int j = 0; j <= 9; ++j)
57 l[i][j] = max(j ? l[i][j - 1] : 0, (i ? l[i - 1][j] : 0) + (j == s[i]));
58 for(int i = n - 1; i >= 0; --i)
59 for(int j = 9; j >= 0; --j)
60 r[i][j] = max(j + 1 <= 9 ? r[i][j + 1] : 0, (i + 1 < n ? r[i + 1][j] : 0) + (j == s[i]));
61 }
62
63 int main() {
64 auto &io = *new IO_<0x10000, 0x10000>;
65
66 int t;
67 io >> t;
68 while(t--) {
69 int n;
70 io >> n >> s;
71 for(int i = 0; i < n; ++i)
72 s[i] -= ‘0‘;
73
74 init_(n);
75
76 int ans = 1, ansl = 0, ansr = 0;
77 for(int a = 0; a <= 9; ++a)
78 for(int b = a; b <= 9; ++b)
79 for(int i = 0; i < n; ++i) {
80 int cl = i ? l[i - 1][a] : 0;
81 int cr = i + 1 < n ? r[i + 1][b] : 0;
82 for(int j = b; j >= a; --j) {
83 f[i][j][0] = ~0x7FFFFFFF, f[i][j][1] = -1; // must not be ans
84 if(j < b)
85 f[i][j][0] = f[i][j + 1][0], f[i][j][1] = f[i][j + 1][1];
86 if(i > 0 && f[i - 1][j][0] + (j == s[i]) > f[i][j][0])
87 f[i][j][0] = f[i - 1][j][0] + (j == s[i]), f[i][j][1] = f[i - 1][j][1];
88 if(j == s[i] && cl + 1 > f[i][j][0])
89 f[i][j][0] = cl + 1, f[i][j][1] = i;
90 int cur = f[i][j][0] + cr;
91 if(cur > ans) {
92 ans = cur;
93 ansl = f[i][j][1];
94 ansr = i;
95 }
96 }
97 }
98 io << ans << ‘ ‘ << ansl + 1 << ‘ ‘ << ansr + 1 << ‘\n‘;
99 }
100
101 delete &io;
102 return 0;
103 }
原文地址：https://www.cnblogs.com/MyGirlfriends/p/9442004.html
时间： 2024-10-07 00:25:51