The 7th Southwest Jiao Tong University Programming Contest—— Critical codes for selected problems

Sel1.

Problem 1708
LinKs: http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1708 

 

 

typedef long long i64d;template<class T> inline T gcd(T a, T b) { if(b == 0) return a; return gcd(b, a % b); }static inline bool chk(i64d x, i64d y){ if(x == y) return 1; static i64d d; d = gcd(x, y); if(d == 1LL) return 0; static i64d _x, _y; _x = x / d, _y = y / d; //_x /= gcd(_x, d), _y /= gcd(_y, d); // 要除到底 static i64d tmp; while( (tmp = gcd(_x, d)) != 1LL ) _x /= tmp; while( (tmp = gcd(_y, d)) != 1LL ) _y /= tmp; return chk(_x, _y);} 

 

Sel2.

Problem 1750
Links: http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1750

 

const int maxV = 1024;const int maxE = 300000;struct node{int v, w;node *nxt;};node *G[maxV];node E[maxE]; int nE = 0;void init(int nV){nE = 0;for(int i = 0; i < nV; ++i) G[i] = 0;}void add(int u, int v, int w){E[nE].v = v, E[nE].w = w;E[nE].nxt = G[u]; G[u] = &E[nE++];}bool find(int &curV, bool tag[], int match[]){for(node *e = G[curV]; e; e = e->nxt) {if( !tag[e->v] ) {tag[e->v] = 1;if( match[e->v] == -1 || find(match[e->v], tag, match) ) {match[e->v] = curV;return 1;}}}return 0;}int maxMatch(int N, int M){int ans = 0;static int match[maxV];memset(match, -1, sizeof(match));static bool tag[maxV];for(int i = 0; i < N; ++i) {memset(tag, 0, sizeof(tag));if( find(i, tag, match) ) ++ans;}return ans;}//add(u - 1, N + v - 1, 1);// int ans = maxMatch(M, M);// printf("Case #%d: %d/n", tt, M - ans); 

 

Sel3.

Problem 1752
Links: http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1752 

int ans = 0, low, high, tmp;for(int i = 0; i < (1 << 2); ++i) {low = inf, high = -inf;for(int j = 0; j < N; ++j) {tmp = 0;if(i & (1 << 0)) tmp += x[j]; else tmp -= x[j];if(i & (1 << 1)) tmp += y[j]; else tmp -= y[j];//cMin(low, tmp);//cMax(high, tmp);if(low > tmp) low = tmp;if(high < tmp) high = tmp;}//cMax(ans, high - low);if(ans < high - low) ans = high - low;}printf("Case #%d: %d/n", tt, ans); 

 

Sel4.

Problem 1757
Links: http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1757

/* * Author: hzwu * Time: 18/04/2011 * Problem: Maximum Sum.cpp */#include<cstdio>#include<cstring>#include<cstdlib>#include<ctime>#include<iostream>#include<algorithm>using namespace std;const int maxN = 100000 + 2;struct elem{int w, idx;elem(){}elem(const int & _w, const int &_idx) : w(_w), idx(_idx) {}inline bool operator<(const elem &_e) const { return w < _e.w; }};struct Heap{private:elem h[maxN << 1]; int nh, i;public:Heap() { nh = 0; }inline void clear() { nh = 0; }inline bool empty() { return nh == 0; }inline elem front() { return h[1]; }inline void push(elem &e) {for(i = ++nh; i > 1 && (e < h[i >> 1]); i >>= 1) {h[i] = h[i >> 1];}h[i] = e;}inline void pop() {elem e = h[nh];for(i = 2; i < nh ; i <<= 1) {if( (i + 1 < nh) && (h[i + 1] < h[i])) ++i;if(e < h[i]) break;h[i >> 1] = h[i];}h[i >> 1] = h[nh--];}} H;int nA, K;int A[maxN];int S[maxN];elem e;int main(){//time_t ss, ee;//ss = clock();freopen("data2.in", "r", stdin);freopen("data2.out", "w", stdout);int t, tt;scanf("%d", &t);for(tt = 1; tt <= t; ++tt) {scanf("%d %d", &nA, &K);for(int i = 0; i < nA; ++i) {scanf("%d", A + i);if(i == 0) S[i] = A[i];else S[i] = S[i-1] + A[i];}for(int i = 0; i < nA; ++i) S[i] -= A[i];//for(int i = 0; i < nA; ++i) printf("%4d", S[i]); puts("");int ans = -1000 * maxN;H.clear();for(int i = 0; i < nA; ++i) {ans = max(ans, A[i]);if( !H.empty() ) {e = H.front();while( (i - e.idx + 1) > K && !H.empty() ) {H.pop(); e = H.front();}//printf("idx = %d, w = %d/n", e.idx, e.w);ans = max(ans, S[i] - e.w + A[i]);//printf("%d->%d/n", e.idx, i);//printf("ans = %d/n", ans);}e.idx = i, e.w = S[i];H.push(e);//printf("ans = %d/n", ans);}printf("Case #%d: %d/n", tt, ans);}//ee = clock();//printf("time = %.2lfms/n", (double)(ee - ss));return 0;} 

 

Sel5.

Problem 1760
Links: http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1760

/** Author: hzwu* Time: 13/03/2011* Problem: Higher Mathematics.cpp*/#include<cstdio>#include<cstring>#include<cmath>#include<ctime> static inline double Area(int C){ return C * log(0.0 + C) - C + 1.0;} static inline int f(int &C){ int ans = 0; for(int i = 1; i <= C; ++i) ans += C / i; return ans;} static inline int get(int &N){ static int low, mid, high, curVal; low = 1, high = N; while(low < high) { mid = (low + high) >> 1; curVal = f(mid); if(curVal >= N) high = mid; else low = mid + 1; } return high;} int main(){//time_t s, e;//s = clock();freopen("in", "r", stdin);freopen("out", "w", stdout); static int testCases, tt = 0; static int N; for(scanf("%d", &testCases); testCases > 0; --testCases) { scanf("%d", &N); printf("Case #%d: %.4lf/n", ++tt, Area( get(N) ) ); }//e = clock();//printf("time = %.2lf/n", (double)(e - s)); return 0;} 

 

Sel6.

Problem 1763
Links: http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1763

/* * Author: hzwu * Time: 25/03/2011 * Problem: Beautiful Probability.cpp */#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const double eps = 1e-9;// if x > 0 , x < 0 or x = 0template<class T> inline int O_O(T x) { return (x > eps) - (x < -eps); }double Lx, Ux, Ly, Uy, Lz, Uz, V;inline double calc(){// x: [Lx, Ux], y: [Ly, Uy], z: [Lz, Uz]// x + y + z <= V// assume x = _x + Lx, y = _y + Ly, z = _z + Lz// then _x: [0, Ux - Lx], _y: [0, Uy - Ly], _z: [0, Uz - Lz]// _x + _y + _z <= V - (Lx + Ly + Lz)// so we can get the follwing equation// _x + _y + _z <= T, where T = V - (Lx + Ly + Lz)// _x: [0, a], _y: [0, b], _z: [0, c]// i think you know values of a, b and c that i can say nothing!static double a, b, c;a = Ux - Lx, b = Uy - Ly, c = Uz - Lz; // a, b and c >= 0double T = V - (Lx + Ly + Lz);if(T + eps < 0) return 0.0; // T < 0if(T > a + b + c - eps) return 1.0; // T >= a + b + cstatic double tmp; // make sure that a <= b <= cif(a > b) { tmp = a; a = b; b = tmp; }if(a > c) { tmp = a; a = c; c = tmp; }if(b > c) { tmp = b; b = c; c = tmp; }if(!O_O(a) && !O_O(b) && !O_O(c)) return 1.0; // a = 0, b = 0 and c = 0if(!O_O(a) && !O_O(b)) return min(1.0, T / c); // a = 0 and b = 0if(!O_O(a)) {// _y + _z <= T/*|(0,T)|.| .| .| .|________._____| (T,0)|*/if(T < b + eps) return (0.5 * T * T) / (b * c);if(T < c + eps) return 0.5 * b * (T + T - b) / (b * c);if(T < b + c + eps) return 1.0 - 0.5 * (b + c - T) * (b + c - T)/ (b * c);return 1.0; // b + c <= T}// T <= a, where we can know that T <= b and T <= cif(T < a + eps) return T * T * T / 6.0 / (a * b * c);// T <= b, where we can know that T <= cif(T < b + eps) return (T * T * T - (T - a) * (T - a) * (T - a)) / 6.0 / (a * b * c);// T <= c// here we should think about T <= a + b and T > a + bif(T < c + eps && T < a + b + eps)return (T * T * T - (T - a) * (T - a) * (T - a) - (T - b) * (T - b) * (T - b)) / 6.0 / (a * b * c);if(T < c + eps && !(T < a + b + eps)) // need to checkreturn (T * T * T - (T - a) * (T - a) * (T - a) - (T - b) * (T - b) * (T - b) + (T - a - b) * (T - a - b) * (T - a - b)) / 6.0 / (a * b * c);if(!(T < c + eps) && T < a + b + eps)return (T * T * T - (T - a) * (T - a) * (T - a) - (T - b) * (T - b) * (T - b) - (T - c) * (T - c) * (T - c)) / 6.0 / (a * b * c);// at this step, we know that T > c and T > a + b and T < a + b + c// but we didn't know does T > c + a and T > c + b// we know that p(_x + _y + _z <= T) = 1 - p(_x + _y + _z > T)// p(_x + _y + _z > T) => p(a - _x_ + b - _y_ + c - _z_ > T)// => p(_x_ + _y_ + _z_ < _T), where _T = a + b + c - T// and _x_, _y_ and _z_ have the same ranges to _x, _y, _z// isn't it?static double _T; _T = a + b + c - T;// p(_x + _y + _z <= T) = 1 - p(_x_ + _y_ + _z_ < _T)// p(_x_ + _y_ + _z_ = _T) ---> 0// p(_x + _y + _z <= T) = 1 - p(_x_ + _y_ + _z_ <= _T)// _T > 0if(_T < a + eps) return 1 - _T * _T * _T / 6.0 / (a * b * c);if(_T < b + eps) return 1 - (_T * _T * _T - (_T - a) * (_T - a) * (_T - a)) / 6.0 / (a * b * c);if(_T < c + eps && _T < a + b + eps)return 1 - (_T * _T * _T - (_T - a) * (_T - a) * (_T - a) - (_T - b) * (_T - b) * (_T - b)) / 6.0 / (a * b * c);if(_T < c + eps && !(_T < a + b + eps)) // need to checkreturn 1 - (_T * _T * _T - (_T - a) * (_T - a) * (_T - a) - (_T - b) * (_T - b) * (_T - b) + (_T - a - b) * (_T - a - b) * (_T - a - b)) / 6.0 / (a * b * c);if(!(_T < c + eps) && _T < a + b + eps)return 1 - (_T * _T * _T - (_T - a) * (_T - a) * (_T - a) - (_T - b) * (_T - b) * (_T - b) - (_T - c) * (_T - c) * (_T - c)) / 6.0 / (a * b * c);// _T must be less than a + b as _T = a + b + (c - T) < a + breturn 1.0;}int main(){//freopen("data.in", "r", stdin);//freopen("data.out", "w", stdout);double ans; static int idx = 0, t;scanf("%d", &t); while( (t--) > 0) {scanf("%lf %lf %lf %lf %lf %lf %lf", &Lx, &Ux, &Ly, &Uy, &Lz, &Uz, &V);ans = calc();printf("Case #%d: %.4lf/n", ++idx, ans);}return 0;} 

 

Sel7.

Problem 1766
Links: http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1766

/* * Author: hzwu * Time: 15/04/2011 * Problem: Stupid Game.cpp */#include<cstdio>#include<cstring>#include<queue>#include<iostream>#include<algorithm>using namespace std;const int maxN = 30;const int inf = 0x7fffffff;struct node {int v, w, idx;node *nxt;};node *G[maxN * maxN];node E[maxN * maxN * maxN * maxN * 3];int nE, s, t;int N, M, v[30][30];void init(int nV){nE = 0;for(int i = 0; i < nV; ++i) G[i] = 0;}void add(int u, int v, int w){E[nE].v = v; E[nE].w = w; E[nE].idx = nE;E[nE].nxt = G[u]; G[u] = &E[nE++];E[nE].v = u; E[nE].w = 0; E[nE].idx = nE;E[nE].nxt = G[v]; G[v] = &E[nE++];}int dir[4][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0} };#define ok(x,y) ((x) >= 0 && (x) < N && (y) >= 0 && (y) < M)int h[maxN * maxN];bool bfs(){memset(h, -1, sizeof(h));queue<int> q; int u;q.push(s); h[s] = 0;while( !q.empty() ) {u = q.front();q.pop();for(node *e = G[u]; e; e = e->nxt) {if( e->w > 0 && h[e->v] == -1 ) {h[e->v] = h[u] + 1;q.push(e->v);}}}return h[t] != -1;}int dfs(int curV, int curF){if(curV == t) return curF;int cntf = 0, tmpf;for(node *e = G[curV]; e; e = e->nxt) {if( (e->w > 0) && (h[e->v] == h[curV] + 1) && cntf < curF && ((tmpf = dfs(e->v, min(e->w, curF - cntf))) > 0) ) {E[e->idx].w -= tmpf;E[(e->idx) ^ 1].w += tmpf;cntf += tmpf;}}if(cntf == 0) h[curV] = -1;return cntf;}int main(){freopen("data.in", "r", stdin);freopen("data.out", "w", stdout);int _t, _tt = 1;scanf("%d", &_t);while( _tt <= _t ) {scanf("%d %d", &N, &M);for(int i = 0; i < N; ++i) {for(int j = 0; j < M; ++j) {scanf("%d", &v[i][j]);}}s = N * M; t = s + 1;init( t + 1 );for(int i = 0; i < N; ++i) {for(int j = 0; j < M; ++j) {if( (i + j) & 1 ) add(s, i * M + j, v[i][j]);else add(i * M + j, t, v[i][j]);}}int tmpx, tmpy;for(int i = 0; i < N; ++i) {for(int j = 0; j < M; ++j) {//printf("[%d] ", i * M + j);if( (i + j) & 1)for(int k = 0; k < 4; ++k) {tmpx = i + dir[k][0];tmpy = j + dir[k][1];if( ok(tmpx, tmpy) ) {add(i * M + j, tmpx * M + tmpy, inf);//printf(" ==> %d", tmpx * M + tmpy);}}//puts("");}}/*for(int i = 0; i <= t; ++i) {printf("[%d]: ", i);for(node *e = G[i]; e; e = e->nxt) {printf("-> %d(%d)", e->v, e->w);}puts("");}*/int ans = 0; for(int i = 0; i < N; ++i) {for(int j = 0; j < M; ++j) {ans += v[i][j];}}//printf("ans = %d/n", ans);while( bfs() ) {ans -= dfs(s, inf);//printf("ans = %d/n", ans);}printf("Case #%d: %d/n", _tt++, ans);}return 0;} 

 

Sel8.

Problem 1767
Links: http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1767

/* * Author: hzwu* Time: 11/03/2011* Problem: Translator.cpp*/#include<cstdio>#include<cstdlib>#include<ctime>#include<map>#include<string>#include<algorithm>#include<iostream>using namespace std;map<int, string> g;void init() {g.clear();g[1] = "one"; g[2] = "two"; g[3] = "three"; g[4] = "four"; g[5] = "five"; g[6] = "six"; g[7] = "seven"; g[8] = "eight"; g[9] = "nine"; g[10] = "ten";g[11] = "eleven"; g[12] = "twelve"; g[13] = "thirteen"; g[14] = "fourteen"; g[15] = "fifteen";g[16] = "sixteen"; g[17] = "seventeen"; g[18] = "eighteen"; g[19] = "nineteen";g[20] = "twenty"; g[30] = "thirty"; g[40] = "forty"; g[50] = "fifty";g[60] = "sixty"; g[70] = "seventy"; g[80] = "eighty"; g[90] = "ninety";g[100] = "one hundred"; g[200] = "two hundred"; g[300] = "three hundred";g[400] = "four hundred"; g[500] = "five hundred"; g[600] = "six hundred";g[700] = "seven hundred"; g[800] = "eight hundred"; g[900] = "nine hundred";for(int i = 21; i <= 99; ++i) {if(i % 10) {g[i] = g[i / 10 * 10] + "-" + g[i % 10];}}for(int i = 101; i <= 999; ++i) {if(i % 100) {g[i] = g[i / 100 * 100] + " and " + g[i % 100];}}}void solve(int n) {if(n == 0) { cout<< "zero" << endl; return; }int a = n / 1000000000;int b = n % 1000000000 / 1000000;int c = n % 1000000 / 1000;int d = n % 1000;if(a) cout << g[a] << " billion";if(b) { if(a) cout<< ", "; cout << g[b] << " million"; }if(c) { if(a||b)cout<< ", "; cout << g[c] << " thousand"; }if(d) { if(a||b||c) cout << ", "; cout << g[d] ;}cout << endl;}int main() {//freopen("Translator.in", "r", stdin);//freopen("Translator.out", "w", stdout);init();int t, n; cin>> t;for(int C = 1; C <= t; ++C) {cin>> n;cout<< "Case #"<< C << ": "; solve(n); }return 0;} 

 

Sel9.

Problem 1768
Links: http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1768

/* * Author: hzwu * Time: 21/04/2011 * Problem: A Simple Problem.cpp */#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long i64d;const int maxM = 100 + 2;const i64d p = 1000000007LL;i64d C[maxM][maxM];void init(){C[0][0] = 1LL % p;C[1][0] = 1LL % p; C[1][1] = 1LL % p;// C[i][j] = C[i - 1][j] + C[i - 1][j - 1];for(int i = 2; i <= 101; ++i) {for(int j = 0; j <= i; ++j) {if( j == 0 || j == i ) C[i][j] = 1LL % p;else if(j > i - j) C[i][j] = C[i][i - j];else C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % p;}}}// calc a ^ b mod pi64d exp_mod(i64d a, i64d b, i64d c){if(b == 0LL) return 1LL % c;if(b == 1LL) return a % c;i64d ret = exp_mod(a, b / 2LL, c);ret = ret * ret % c;if(b % 2LL == 1LL) ret = ret * a % c;return ret;}i64d F[maxM]; // record F[N][j], j = 0...M// F[maxM]void calc_F(i64d N){F[0] = ((N - 1LL) % p + p) % p;for(int i = 2; i <= 101; ++i) {F[i - 1] = exp_mod(N + 1LL, i, p);F[i - 1] = F[i - 1] - exp_mod(2LL, (i64d)i, p) % p;F[i - 1] = ( F[i - 1] + p ) % p;for(int j = 0; j <= i - 2; ++j) {F[i - 1] = F[i - 1] - C[i][j] * F[j] % p;F[i - 1] = ( F[i - 1] + p ) % p;}F[i - 1] = (F[i - 1] * exp_mod((i64d)i, p - 2LL, p)) % p;}}i64d calc(i64d A, i64d D, i64d M, i64d N){if( N == 0LL ) return exp_mod(A, M, p);if( N == 1LL ) return (exp_mod(A, M, p) + exp_mod(A + D, M, p)) % p;calc_F(N); // calc F[maxM]i64d ans = (exp_mod(A, M, p) + exp_mod(A + D, M, p)) % p, tmp;// sum { (A + K * D) ^ M | 2 <= K <= N }// = sum { C[M][i] * (A ^ (M - i)) * (D ^ i) * F[i] | 0 <= i <= M }for(i64d i = 0; i <= M; ++i) {tmp = ((C[M][i] * (exp_mod(A, M - i, p) % p) % p) * exp_mod(D, i, p)) % p * F[i] % p;ans = (ans + tmp) % p;}return ans;}i64d A, D, M, N;int main(){freopen("data.in", "r", stdin);freopen("data.out", "w", stdout);init(); // tableint t, tt;scanf("%d", &t);for(tt = 1; tt <= t; ++tt) {scanf("%lld %lld %lld %lld", &A, &D, &M, &N);printf("Case #%d: %lld/n", tt, calc(A, D, M, N));}return 0;} 

 

Sel10.

Problem 1770
Links: http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1770

/* * Author: hzwu * Time: 03/05/2011 * Problem: Next Generation.cpp */#include<cstdio>#include<cstring>const int mod = 2021;const int maxN = 80 + 1;const int maxM = 10000 + 1;static int dp[maxN][maxM];static int N, M;void init() {memset(dp, 0, sizeof(dp));dp[1][0] = 1;for(int i = 2; i <= 80; ++i) {for(int j = 1; j <= 10000; ++j) {dp[i][j] += 2 * dp[i-1][j-1];if(j > 1) dp[i][j] += (i - 2 ) * dp[i-1][j-2];dp[i][j] %= mod;}}}int main() {freopen("data.in", "r", stdin);freopen("data.out", "w", stdout);init();int t, tt = 1;scanf("%d", &t);while( (t --) > 0 ) {scanf("%d%d", &N, &M);printf("Case #%d: %d/n", tt++, dp[N][M]);}return 0;}/*#include<cstdio>#include<cstring>#include<cstdlib>#include<ctime>int main(){freopen("data.in", "w", stdout);srand((unsigned)time(NULL));puts("10085");int N, M;for(int i = 0; i < 5000; ++i) {N = rand() % 80 + 1;M = rand() % N;printf("%d %d/n", N, M);}for(int i = 5000; i < 10085; ++i) {N = rand() % 80 + 1;M = rand() % 10001;printf("%d %d/n", N, M);}return 0;}*/ 

 

Sel11.

Problem 1771
Links: http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1771

/* * Author: hzwu * Time: 06/05/2011 * Problem: Rich Obama.cpp */// dp[i][j + v] = min(dp[i][j + v], dp[i][j] + 1)// ==>// dp[j + v] = min(dp[j + v], dp[j] + 1)#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int maxN = 10000 + 2;static int dp[maxN << 1];int totV, N, v;int main(){ //freopen("data3.in", "r", stdin); //freopen("data3.out", "w", stdout); int t, tt; scanf("%d", &t); for(tt = 1; tt <= t; ++tt) { scanf("%d %d", &totV, &N); memset(dp, 0x3f, sizeof(dp)); dp[0] = 0; while( (N --) > 0 ) { scanf("%d", &v); for(int _v = totV; _v >= 0; --_v) { if(dp[_v + v] > dp[_v] + 1) { dp[_v + v] = dp[_v] + 1; } } } for(int curV = totV; curV < (maxN << 1); ++curV) { if( dp[curV] < 0x3f3f3f3f ) { printf("Case #%d: %d %d/n", tt, curV, dp[curV]); break; } } } return 0;} 

 

NOTES: Please do not copy code to submit which is a stupid action!

 

by hzwu.