西山居初赛二 总结
hdu 4548 美素数
用线性筛法筛选出10^6的素数，然后预处理统计下.. O(1)就能得到结果了.
#include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> using namespace std; const int N = (int)1e6+10; bool vis[N]; int p[N], sum[N]; void GetPrime(){ memset(vis,0,sizeof(0)); vis[1] = 1; int cnt = 0; for(int i = 2; i < N; i++){ if( !vis[i] ) p[cnt++] = i; for(int j = 0; (j<cnt)&&(p[j]*i<N); j++){ vis[ p[j]*i ] = 1; if( i%p[j] == 0 )break; } } } int Dight(int x){ int s = 0; while(x) s+=x%10, x /= 10; return s; } void init(){ GetPrime(); sum[0] = sum[1] = 0; for(int i = 2; i < N; i++){ sum[i] = sum[i-1]; if( !vis[i] && !vis[ Dight(i) ] ) sum[i]++; } } int main(){ init(); int T; scanf("%d", &T); for(int Case = 1; Case <= T; Case++ ){ int a, b; scanf("%d%d", &a,&b); printf("Case #%d: %d\n", Case, sum[b] - sum[a-1] ); } return 0; }
hdu 4549 M斐波那契数列
直接搞显然不行, 没有线性关系. 简单推导下有, a, b, a^1*b^1, a^1*b^2 .... 可以知道a,b的幂满足Fib, 然后构造矩阵快速幂...就好了.还需要个性质: A^X = A^( X mod Eular(M) ) ( mod M ) .
#include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; typedef long long LL; const int mod = (int)1e9+7; struct Matrix{ LL mat[2][2]; void zero(){memset(mat,0,sizeof(mat)); } void unit(){zero();mat[0][0]=mat[1][1]=1;} }A, T; void init_Matrix(){ A.zero(); T.zero(); A.mat[0][0]=A.mat[1][0]=1; T.mat[0][1]=T.mat[1][0]=T.mat[1][1]=1; } void print(Matrix a){ printf("Matrix:\n"); for(int i = 0; i < 2; i++){ for(int j = 0; j < 2; j++) printf("%d ", (int)a.mat[i][j]); puts(""); } } Matrix mult(Matrix a, Matrix b){ Matrix c; c.zero(); for(int i = 0; i < 2; i++) for(int j = 0; j < 2; j++) for(int k = 0; k < 2; k++) c.mat[i][k]=(c.mat[i][k]+a.mat[i][j]*b.mat[j][k])%(mod-1); return c; } Matrix Pow(Matrix x,int n){ Matrix c; c.unit(); while( n ){ if(n&1) c = mult(x,c); x = mult(x,x); n >>= 1; } return c; } LL Fib(int n){ init_Matrix(); A = mult( Pow(T,n-1), A ); return A.mat[0][0]; } LL pow( LL x, int n ){ LL c = 1; while(n){ if(n&1) c = c*x%mod; x = x*x%mod; n >>= 1; } return c; } int main(){ int a, b, n; while( scanf("%d%d%d",&a,&b,&n) != EOF){ if( n < 2 ){ if( n == 0 ) printf("%d\n", a); else printf("%d\n", b); } else{ LL n1 = Fib( n-1 ), n2 = Fib( n ); // printf("n1 = %d, n2 = %d\n", (int)n1, (int)n2 ); LL ans = pow(a,n1)*pow(b,n2)%mod; printf("%d\n", (int)ans ); } } return 0; }
hdu 4550卡片游戏
和波神讨论了好久...有0的情况好恶心. 最后邪恶的处理第一次出现0 的情况. 主要是出现了0, 则此时,在之后选定一个最小非0的数放在前面必定最优. 其他情况就是大的放后,小的放前面.....稀里哗啦写了好多.. A掉了后看到别人代码好短- - .....才发现好多地方做了无用工.......
#include<cstdio> #include<cstring> #include<cstdlib> #include<string> #include<algorithm> using namespace std; const int N = 110; char s[N]; string t; int len; // S( i, len-1 ) void gao( int i ){ char tm = '9'+10; int pos = -1, cnt = 0; for(int j = i; j < len; j++){ if( (s[j]!='0') && (tm>=s[j]) ) tm = s[j], pos = j; } if( (pos==-1) || ( (t[0]!='0')&&(s[pos]>t[0]))){ for(int j = i; j < len; j++) t = t+s[j]; } else{ for(int j = i; j < pos; j++){ if( s[j] == '0' ) cnt++; else t = t+s[j]; } for(int j = 0; j < cnt; j++ ) t = '0' + t; t = s[pos] + t; for(int j = pos+1; j < len; j++ ) t = t + s[j]; } //printf("%s\n", t.c_str() ); } int main(){ int T; scanf("%d", &T); while( T-- ){ scanf("%s", s); t.clear(); len = strlen(s); int i = 1; t = s[0]; if( s[0] == '0' ) gao(1); else{ while( (i<len) && (s[i]!='0') ){ if( s[i] > t[0] ) t = t+s[i]; else t = s[i] + t; i++; } if( i < len ) gao(i); } printf("%s\n", t.c_str() ); } return 0; }
精简的代码思路: 就是找一个非0最小所在下标pos, 放在最前面. 其中 下标i < pos的, 用大在后,小在前. 然后i > pos的直接原样丢后面...不用考虑0的特殊情况....真心好..
#include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> using namespace std; const int N = 110; char s[N]; int main(){ int _; scanf("%d", &_); while( _-- ){ scanf("%s", s); string t; t += s[0]; char ch = '9'; int pos; for(int i = 0; s[i]; i++) if( (s[i]!='0') && (ch>=s[i]) ) ch = s[pos=i]; for(int i = 1; s[i]; i++){ if( i == pos ) t = s[i] + t; else if( i < pos ){ if( s[i] <= t[0] ) t = s[i] + t; else t = t + s[i]; } else t = t + s[i]; } printf("%s\n", t.c_str() ); } return 0; }