2015 上海网络赛 ACM/ICPC Asia Regional Shanghai Online

An easy problem

题意: 两种操作 1 乘一个数 2 除一个之前乘过的数  输出答案模M

分析: 线段树 - - 每次操作更新单点 然后求区间乘积 答案就是mul[1]

代码:

////  Created by TaoSama on 2015-09-26//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;int n, M;int mul[N << 2];#define root 1, n, 1#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1void build(int l, int r, int rt) {    mul[rt] = 1;    if(l == r) return;    int m = l + r >> 1;    build(lson);    build(rson);}void update(int o, int v, int l, int r, int rt) {    if(l == r) {        mul[rt] = v % M;        return;    }    int m = l + r >> 1;    if(o <= m) update(o, v, lson);    else update(o, v, rson);    mul[rt] = 1LL * mul[rt << 1] * mul[rt << 1 | 1] % M;}int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d%d", &n, &M);        build(root);        printf("Case #%d:\n", ++kase);        for(int i = 1; i <= n; ++i) {            int op, x; scanf("%d%d", &op, &x);            if(op == 1) update(i, x, root);            else update(x, 1, root);            printf("%d\n", mul[1]);        }    }    return 0;}

MZ大神的思路:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>using namespace std ;typedef long long ll ;const int maxn = 1e5 + 5 ;const int inf = 0x3f3f3f3f  ;const ll mod = 1e9 + 7 ;const double eps = 1e-6 ;const double pi = acos(-1.0) ;int t , n , m , kase ;ll a[maxn] ;int o , q ;int p[20] , c , cnt[20] ;void fac( int num ){    c = 0 ;    for( int i = 2 ; i*i <= num ; i++ )    {        if( num % i == 0 )        {            p[c++] = i ;            while( num % i == 0 ) num /= i ;        }    }    if( num != 1 ) p[c++] = num ;}void update( int &num , int k ){    for( int i = 0 ; i < c ; i++ )    {        while( num % p[i] == 0 )        {            num /= p[i] ;            cnt[i] += k ;        }    }}void show(){    for( int i = 0 ; i < c ; i++ ) printf( " p[%d] = %d\n" , p[i] , cnt[i] ) ;}ll quickPow( ll a , ll b , ll c ){    ll r = 1 ;    while( b > 0 )    {        if( b&1 ) r = r*a%m ;        b >>= 1 ;        a = a*a%m ;    }    return r ;}void exgcd( ll a , ll b , ll &d , ll &x , ll &y ){    if( !b ) { d = a , x = 1 , y = 0 ; }    else    {        exgcd(b,a%b,d,y,x) ;        y -= x*(a/b) ;    }}ll inv( ll num ){    ll x , y , d ;    exgcd(num,m,d,x,y) ;    return ( x%m + m ) % m ;}ll get( ){    ll ret = 1 ;    for( int i = 0 ; i < c ; i++ ) ret = ret * quickPow( p[i] , cnt[i] , m ) % m ;    return ret ;}int main(){    scanf( "%d" , &t ) ;    while( t-- )    {        scanf( "%d%d" , &q , &m ) ;        fac(m) ;        memset( cnt , 0 , sizeof cnt ) ;        ll ans = 1 ;        printf( "Case #%d:\n" , ++kase ) ;        for( int i = 1 ; i <= q ; i++ )        {            scanf( "%d%I64d" , &o , a+i ) ;            if( o == 1 )            {                int tp = a[i] ;                update( tp , 1 ) ;                ans = ans * tp % m ;            }            else            {                int tp = a[a[i]] ;                update( tp , -1 ) ;                ans = ans * inv(tp) % m ;            }            printf( "%I64d\n" , ans*get()%m ) ;        }    }    return 0 ;}

Explore Track of Point

分析: http://blog.csdn.net/Baileys0530/article/details/48753151

          或者延长高线 向两腰做垂线 画圆弧
代码:

////  Created by TaoSama on 2015-09-26//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;const double PI = acos(-1);struct Point {    double x, y;    void read() {        scanf("%lf%lf", &x, &y);    }} A, B, C, M;double getdis(Point& a, Point& b) {    return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2));}int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        A.read(), B.read(), C.read();        M.x = (B.x + C.x) / 2, M.y = (B.y + C.y) / 2;        double a = getdis(B, M), b = getdis(A, B), h = getdis(A, M);        double r = 2 * a * h / (2 * a + 2 * b);  //内接圆半径        double R = (a * a - r * r) / r / 2 + r; //        double ans = 2 * R * asin(a / R) + h;        printf("Case #%d: %.4f\n", ++kase, ans);    }    return 0;}

Can you find it

分析: http://blog.csdn.net/queuelovestack/article/details/48754331

          快速幂的时候可以用  降幂公式降幂 - - 素数就是 (C-1)

代码:

////  Created by TaoSama on 2015-09-26//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;LL C, k1, b1, k2;LL ksm(LL x, LL n, LL m) {    LL ret = 1;    while(n) {        if(n & 1) ret = ret * x % m;        x = x * x % m;        n >>= 1;    }    return ret;}int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int kase = 0;    while(scanf("%d%d%d%d", &C, &k1, &b1, &k2) == 4) {        bool ok = false;        printf("Case #%d:\n", ++kase);        for(int a = 1; a < C; ++a) {            int b = C - ksm(a, (k1 + b1) % (C - 1), C);            if(ksm(a, k1 % (C - 1), C) == ksm(b, k2 % (C - 1), C)) {                ok = true;                printf("%d %d\n", a, b);            }        }        if(!ok) puts("-1");    }    return 0;}