Codeforces Round #373 (Div. 2)

Posted 2020-08-09

岑寂了一个暑假之后终于涨分了，借助了A题的hack大法， c题没有理解好题意，wa就放弃了，后来再样例的帮助下。。终于AC

现在还没搞懂矩阵啊。烦。。抄来别人的代码

A. Vitya in the Countryside

当n = 1时候特判0，15

 1 #include <iostream>
 2 #include <cstdio>
 3 
 4 using namespace std;
 5 int main() {
 6     int n, m;
 7     int a[100];
 8     scanf("%d", &n);
 9     for(int i = 0;i < n; i++){
10         scanf("%d", &a[i]);
11     }
12     if(a[n-1] == 15) printf("DOWN\n");
13     else if(a[n-1] == 0) printf("UP\n");
14     else if(n == 1) printf("-1\n");
15     else if(a[n-1] < a[n-2]){
16         printf("DOWN\n");
17     }else printf("UP\n");
18     return 0;
19 }

B. Anatoly and Cockroaches

分别计算奇偶位r、b的个数，求一下就好

 1 #include <iostream>
 2 #include <cstdio>
 3 
 4 using namespace std;
 5 int main() {
 6     int n, m;
 7     char s[100002];
 8     scanf("%d", &n);
 9     scanf("%s", s);
10     int first_r = 0, first_b = 0, second_r = 0, second_b = 0;
11     for(int i = 0; i < n; i++){
12         if(i%2 == 0) {
13             if(s[i] == ‘r‘) first_r++;
14             else first_b++;
15         }else {
16             if(s[i] == ‘b‘) second_b++;
17             else second_r++;
18         }
19     }
20     //printf("%d %d %d %d\n", first_r, first_b, second_r, second_b);
21     int ans1 = first_b + second_r - min(first_b, second_r);
22     int ans2 = first_r + second_b - min(first_r, second_b);
23     printf("%d\n", min(ans1, ans2));
24     return 0;
25 }

C. Efim and Strange Grade

搞得我也很恶心了， 分类讨论各个情况

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int main() {
 6     int n, m;
 7     scanf("%d%d", &n, &m);
 8     char s[200005];
 9     scanf("%s", s);
10     int flag = 0, flags = 1;
11     int ans = -1;
12     for(int i = 0; i < n; i++) {
13         if(s[i] == ‘.‘) {
14             flag = 1;
15             continue;
16         }
17         if(s[i] - ‘0‘ >= 5 && flag == 1 && m > 0) {
18             ans = i;
19             m--;
20             break;
21         }
22     }
23     if(ans == -1) printf("%s\n", s);
24     else {
25         int snow = 1;
26         if(s[ans-1] == ‘.‘) ans--, flags = 0;
27         for(int i = ans-1; i >= 0; i--) {
28             if(s[i] == ‘.‘) continue;
29             int sum = s[i] - ‘0‘ + snow;
30             s[i] = sum%10 + ‘0‘;
31             snow = sum/10;
32         }
33         if(snow == 1) {
34             printf("1");
35             for(int i = 0; i < ans; i++) printf("%c", s[i]);
36             printf("\n");
37         } else {
38             int last = ans, now = 0;
39             for(int i = last-1; i >= 0; i--) {
40                 if(s[i] == ‘.‘) {
41                     flags = 0;
42                     ans = i-1;
43                     continue;
44                 }
45                 int sum = s[i] - ‘0‘ + now;
46                // printf("%d %d %d %d\n", i, now, sum, flags);
47                 if((sum <= 4 || m <=  0) && flags) {
48                     now = 0;
49                     s[i] = sum + ‘0‘;
50                     ans = i;
51                     break;
52                 } else if(flags) {
53                     m--;
54                     now = 1;
55                     ans--;
56                 } else {
57                     s[i] = sum%10 + ‘0‘;
58                     now = sum /10;
59                 }
60             }
61             if(now == 1) {
62                 printf("1");
63                 for(int i = 0; i <= ans; i++) printf("%c", s[i]);
64                 printf("\n");
65             }
66             else {
67             if(s[ans] < ‘0‘  || s[ans] > ‘9‘) ans--;
68             for(int i = 0; i <= ans; i++) printf("%c", s[i]);
69             printf("\n");
70             }
71         }
72     }
73     return 0;
74 }

E. Sasha and Array（抄的代码）

  1 #include<cstdio>
  2 using namespace std;
  3 typedef long long LL;
  4 const int N = 100010;
  5 const int P = 1000000007;
  6 
  7 struct Matrix {
  8     int n,m;
  9     LL c[3][3];
 10     Matrix() {
 11         for(int i=0; i<2; i++)for(int j=0; j<2; j++)c[i][j]=0;
 12         n=m=0;
 13         return;
 14     }
 15     Matrix operator * (const Matrix & a) const {
 16         Matrix b;
 17         b.n=n;
 18         b.m=a.m;
 19         for(int i=0; i<n; i++)
 20             for(int j=0; j<b.m; j++) {
 21                 for(int k=0; k<m; k++)
 22                     b.c[i][j]=(b.c[i][j]+c[i][k]*a.c[k][j])%P;
 23             }
 24         return b;
 25     }
 26     Matrix operator + (const Matrix & a) const {
 27         Matrix b;
 28         b.n=n;
 29         b.m=m;
 30         for(int i=0; i<n; i++)
 31             for(int j=0; j<m; j++)
 32                 b.c[i][j]=(a.c[i][j]+c[i][j])%P;
 33         return b;
 34     }
 35 } o,ini,oo,T[N<<2],tag[N<<2];
 36 
 37 Matrix FP(Matrix a,int x) {
 38     oo=o;
 39     for(; x; x>>=1,oo=oo*oo)
 40         if(x&1) a=a*oo;
 41     return a;
 42 }
 43 
 44 int n,m,seq[N];
 45 
 46 void Build(int x,int l,int r) {
 47     T[x].n=1,T[x].m=2;
 48     if(l==r) {
 49         tag[x].n=tag[x].m=0;
 50         T[x]=ini;
 51         T[x]=FP(T[x],seq[l]-1);
 52         return;
 53     }
 54     int mid=(l+r)>>1;
 55     Build(x<<1,l,mid);
 56     Build(x<<1|1,mid+1,r);
 57     T[x]=T[x<<1]+T[x<<1|1];
 58     return;
 59 }
 60 
 61 void Pushdown(int x) {
 62     if(tag[x].n==0) return;
 63     T[x<<1]=T[x<<1]*tag[x];
 64     if(tag[x<<1].n==0) tag[x<<1]=tag[x];
 65     else tag[x<<1]=tag[x<<1]*tag[x];
 66     T[x<<1|1]=T[x<<1|1]*tag[x];
 67     if(tag[x<<1|1].n==0) tag[x<<1|1]=tag[x];
 68     else tag[x<<1|1]=tag[x<<1|1]*tag[x];
 69     tag[x].n=tag[x].m=0;
 70     return;
 71 }
 72 
 73 void Change(int x,int l,int r,int L,int R,Matrix d) {
 74     if(L<=l&&r<=R) {
 75         if(tag[x].n==0) tag[x]=d;
 76         else tag[x]=tag[x]*d;
 77         T[x]=T[x]*d;
 78         return;
 79     }
 80     Pushdown(x);
 81     int mid=(l+r)>>1;
 82     if(L<=mid) Change(x<<1,l,mid,L,R,d);
 83     if(R>mid) Change(x<<1|1,mid+1,r,L,R,d);
 84     T[x]=T[x<<1]+T[x<<1|1];
 85     return;
 86 }
 87 
 88 LL Query(int x,int l,int r,int L,int R) {
 89     if(L<=l&&r<=R) return T[x].c[0][1];
 90     Pushdown(x);
 91     int mid=(l+r)>>1;
 92     LL res=0;
 93     if(L<=mid) res+=Query(x<<1,l,mid,L,R);
 94     if(R>mid) res+=Query(x<<1|1,mid+1,r,L,R);
 95     res%=P;
 96     return res;
 97 }
 98 
 99 int main() {
100     ini.n=1;
101     ini.m=2;
102     ini.c[0][1]=1ll;
103     o.n=2,o.m=2;
104     o.c[1][0]=1ll;
105     o.c[0][1]=1ll;
106     o.c[1][1]=1ll;
107 
108     scanf("%d%d",&n,&m);
109     for(int i=1; i<=n; i++) scanf("%d",&seq[i]);
110     Build(1,1,n);
111 
112     int typ,l,r,x;
113     for(int i=1; i<=m; i++) {
114         scanf("%d%d%d",&typ,&l,&r);
115         if(typ==2) printf("%I64d\n",Query(1,1,n,l,r));
116         else {
117             scanf("%d",&x);
118             Change(1,1,n,l,r,FP(o,x-1));
119         }
120     }
121     return 0;
122 }