第一次比赛的时候离4题最近的一次，可惜思维太慢加手速太慢最终没有能在比赛中A掉。
A：可知，能且只能操作MIN（N，M）次，所以判断奇偶即可
1 // File Name: a.cpp 2 // Author: darkdream 3 // Created Time: 2014年07月24日 星期四 23时31分20秒 4 5 #include<vector> 6 #include<list> 7 #include<map> 8 #include<set> 9 #include<deque> 10 #include<stack> 11 #include<bitset> 12 #include<algorithm> 13 #include<functional> 14 #include<numeric> 15 #include<utility> 16 #include<sstream> 17 #include<iostream> 18 #include<iomanip> 19 #include<cstdio> 20 #include<cmath> 21 #include<cstdlib> 22 #include<cstring> 23 #include<ctime> 24 25 using namespace std; 26 27 int main(){ 28 int n , m ; 29 scanf("%d %d",&n,&m); 30 n = min(n,m); 31 if(n % 2 == 1) 32 printf("Akshat "); 33 else printf("Malvika "); 34 return 0; 35 }
B：分几种情况讨论即可
1 // File Name: b.cpp 2 // Author: darkdream 3 // Created Time: 2014年07月24日 星期四 23时40分41秒 4 5 #include<vector> 6 #include<list> 7 #include<map> 8 #include<set> 9 #include<deque> 10 #include<stack> 11 #include<bitset> 12 #include<algorithm> 13 #include<functional> 14 #include<numeric> 15 #include<utility> 16 #include<sstream> 17 #include<iostream> 18 #include<iomanip> 19 #include<cstdio> 20 #include<cmath> 21 #include<cstdlib> 22 #include<cstring> 23 #include<ctime> 24 25 using namespace std; 26 int a[100005]; 27 int b[100005]; 28 int main(){ 29 int n ; 30 scanf("%d",&n); 31 int ok = 0 ; 32 for(int i = 1;i <= n;i ++) 33 { 34 scanf("%d",&a[i]); 35 if(a[i] < a[i-1]) 36 { 37 b[i] = b[i-1]+1; 38 b[i-1] = 0; 39 } 40 } 41 a[n+1] = 1e9+7; 42 int sum = 0 ; 43 int site = 0 ; 44 for(int i =1 ;i <= n;i ++) 45 { 46 if(b[i]) 47 { 48 sum ++ ; 49 site = i; 50 } 51 } 52 53 if(sum >= 2) 54 printf("no "); 55 if(sum == 0 ) 56 printf("yes 1 1 "); 57 if(sum == 1 ) 58 { 59 if(a[site] > a[site -b[site]-1] && a[site-b[site]] < a[site+1]) 60 printf("yes %d %d ",site-b[site],site); 61 else printf("no "); 62 } 63 return 0; 64 }
C：二进制枚举所有情况，再看其是否满足K和M 的要求
1 // File Name: c.cpp 2 // Author: darkdream 3 // Created Time: 2014年07月24日 星期四 23时57分43秒 4 5 #include<vector> 6 #include<list> 7 #include<map> 8 #include<set> 9 #include<deque> 10 #include<stack> 11 #include<bitset> 12 #include<algorithm> 13 #include<functional> 14 #include<numeric> 15 #include<utility> 16 #include<sstream> 17 #include<iostream> 18 #include<iomanip> 19 #include<cstdio> 20 #include<cmath> 21 #include<cstdlib> 22 #include<cstring> 23 #include<ctime> 24 25 #define LL long long 26 using namespace std; 27 28 int main(){ 29 int ca; 30 scanf("%d",&ca); 31 while(ca--) 32 { 33 LL n , k ,d1,d2; 34 scanf("%I64d %I64d %I64d %I64d",&n,&k,&d1,&d2); 35 LL a[4]; 36 LL b[4]; 37 LL is = 0 ; 38 for(LL i = 0;i <= 3; i ++) 39 { 40 memset(a,0,sizeof(a)); 41 memset(b,0,sizeof(b)); 42 LL t = k ; 43 if(i % 2 == 0 ) 44 { 45 t -= d1; 46 b[1] = d1; 47 } 48 else { 49 t += d1; 50 b[1] = -d1; 51 } 52 53 if(i /2 == 0 ){ 54 t -=d2; 55 b[3] = d2; 56 } 57 else{ t += d2; 58 b[3] = -d2; 59 } 60 if(t % 3 != 0 ) 61 continue; 62 63 LL ok = 0 ; 64 for(LL i = 1;i <= 3 ;i ++) 65 { 66 a[i] = t/3 + b[i]; 67 if(a[i] < 0) 68 ok = 1; 69 } 70 if(ok == 1 ) 71 continue; 72 73 if(n % 3 != 0) 74 continue; 75 LL p = n/3; 76 77 ok = 0 ; 78 for(LL i =1;i <= 3;i ++) 79 { 80 if(a[i]>p) 81 ok = 1; 82 } 83 if(ok == 1) 84 continue; 85 is = 1; 86 break; 87 } 88 if(is) 89 printf("yes "); 90 else printf("no "); 91 92 } 93 return 0; 94 }
D：因为字符串只有A，B组成，可知一个字符串的字串个数有 len*(len+1)/2 种，所以只有两个端点取到相同的字符才能构成 good 字符串，分奇偶的话就只需要统计
每种字符出现的位置有多少个偶数，多少个奇数，排除只有一个字母的字串的情况（单独计数）同为偶数或奇数的长度为奇数，奇偶互异的长度为偶数，想清楚这些就可以求解了。
1 // File Name: d.cpp 2 // Author: darkdream 3 // Created Time: 2014年07月25日 星期五 00时31分33秒 4 5 #include<vector> 6 #include<list> 7 #include<map> 8 #include<set> 9 #include<deque> 10 #include<stack> 11 #include<bitset> 12 #include<algorithm> 13 #include<functional> 14 #include<numeric> 15 #include<utility> 16 #include<sstream> 17 #include<iostream> 18 #include<iomanip> 19 #include<cstdio> 20 #include<cmath> 21 #include<cstdlib> 22 #include<cstring> 23 #include<ctime> 24 #define LL long long 25 using namespace std; 26 char str[100005]; 27 int sum[100005]; 28 LL pike2(LL x) 29 { 30 return x*(x-1)/2; 31 } 32 int main(){ 33 scanf("%s",str); 34 int len = strlen(str); 35 int t= 0; 36 int p = 0 ; 37 for(int i =1 ;i < len ;i ++) 38 { 39 if(str[i] != str[i-1]) 40 { 41 t ++ ; 42 sum[t] = i-p; 43 p = i; 44 } 45 } 46 t ++ ; 47 sum[t] = len -p ; 48 LL sume ,sumo; 49 sume = sumo = 0 ; 50 LL sumek[3]; 51 LL sumok[3]; 52 LL ans = 1ll*len*(len+1)/2; 53 LL tsum = 0 ; 54 memset(sumek,0,sizeof(sumek)); 55 memset(sumok,0,sizeof(sumok)); 56 for(int i =1;i <= t ;i ++) 57 { 58 if(i % 2 == 1) 59 { 60 sumo += sum[i]; 61 if(sum[i] % 2 == 0 ) 62 { 63 sumok[1] += sum[i]/2; 64 sumok[2] += sum[i]/2; 65 }else{ 66 if((tsum+1)%2 == 1) 67 { 68 sumok[1] += sum[i]/2+1; 69 sumok[2] += sum[i]/2; 70 }else{ 71 sumok[1] += sum[i]/2; 72 sumok[2] += sum[i]/2+1; 73 } 74 } 75 76 }else { 77 if(sum[i] % 2 == 0 ) 78 { 79 sumek[1] += sum[i]/2; 80 sumek[2] += sum[i]/2; 81 }else{ 82 if((tsum+1)%2 == 1) 83 { 84 sumek[1] += sum[i]/2+1; 85 sumek[2] += sum[i]/2; 86 }else{ 87 sumek[1] += sum[i]/2; 88 sumek[2] += sum[i]/2+1; 89 } 90 } 91 sume += sum[i] ; 92 } 93 tsum += sum[i]; 94 } 95 // printf("%lld %lld %lld %lld ",sumok[1],sumok[2],sumek[1],sumek[2]); 96 ans = ans - sumo*sume; 97 LL temp = pike2(sumek[1])+pike2(sumek[2])+pike2(sumok[1]) + pike2(sumok[2])+sumo+sume; 98 printf("%lld %lld ",ans-temp,temp); 99 return 0; 100 }
又看到Bman巨简洁的代码，膜拜一下
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6 const int maxn = 1e5 + 10; 7 char s[maxn]; 8 int a[2][2]; 9 int main() 10 { 11 scanf("%s", s); 12 long long x = 0, y = 0; 13 for(int i = 0; s[i] != ' '; i++) 14 { 15 a[i&1][s[i] - 'a']++; 16 x += a[i&1][s[i] - 'a']; 17 y += a[(i&1)^1][s[i] - 'a']; 18 } 19 cout << y << " " << x; 20 return 0; 21 }
E：容斥定理 未解