To be an ACMan

2012年12月7日

失算啊，水题啊，比赛没AView Code #include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[100005], b[100005];int main(){ int i, j, n; scanf("%d", &n); for(i = 0; i < n ;i++) scanf("%d", &a[i]), b[i] = a[i]; if(n <= 2) { puts("-1" Read More

posted @ 2012-12-07 16:17 To be an ACMan Views(321) Comments(0) Diggs(0)

codeforces 251A Points on Line 二分 or 单调队列

比赛时我用单调队列A的View Code #include<cstdio>#include<cstring>#include<queue>using namespace std;#define LL __int64int a, n, d;LL ans;queue <int> q;int main() { int i, j; scanf("%d%d", &n, &d); for(i = 0; i < n; i++) { scanf("%d", &a); q.push(a); wh Read More

posted @ 2012-12-07 15:41 To be an ACMan Views(358) Comments(1) Diggs(0)

2012年12月5日

HDU 1163 Eddy's digital Roots 水题

View Code #include<cstdio>#include<cstring>int n, sum, p;void gao(int &sum, int &p){ while(sum >= 10) { p = 0; while(sum) p += sum % 10, sum /= 10; sum = p; }}int main(){ int i, j; while( ~scanf("%d", &n) && n) { sum = 1; ... Read More

posted @ 2012-12-05 12:42 To be an ACMan Views(136) Comments(0) Diggs(0)

2012年12月4日

HDU 1492 The number of divisors(约数) about Humble Numbers 水题

View Code #include<cstdio> #include<cstring> __int64 n, p, a[4] = {2, 3, 5, 7}; int main() { int i, j; while( ~scanf("%I64d", &n) && n) { __int64 ans = 1; for(i = 0; i < 4; i++) { __int64 m = n; p = 0; ... Read More

posted @ 2012-12-04 23:19 To be an ACMan Views(157) Comments(0) Diggs(0)

HDU 1124 Factorial 水题

题意：统计n！后面有几个0, 2肯定够用，只要算n！里有几个5就可以了View Code #include<cstdio>#include<cstring>int main(){ int cas, n; scanf("%d", &cas); while(cas--) { scanf("%d", &n); int s = 5, ans = 0; while(s <= n) ans += n/s, s *= 5; printf("%d\n", ans); } return 0;} Read More

posted @ 2012-12-04 22:01 To be an ACMan Views(159) Comments(0) Diggs(0)

HDU1214 圆桌会议水规律

分成2段a和b，若n为奇数， a，b一奇一偶，若n为偶数，a，b都为偶数。假设长度为L，每段的次数是 1+2+3+......+L = (L+1)*L/2, a,b两段加起来就是答案;View Code #include<cstdio>#include<cstring>__int64 n, a, b;int main(){ while( ~scanf("%I64d", &n)) a = n>>1, b = n - a, printf("%I64d\n", (a*a+b*b-n)>>1); retu Read More

posted @ 2012-12-04 21:47 To be an ACMan Views(186) Comments(0) Diggs(0)

HDU 1262 寻找素数对素数水题

View Code #include<cstdio>#include<cstring>#include<cmath>bool vis[10004];void init() { int i, j; for(i = 2; i * i <= 10000; i++) for(j = i*i; j <= 10000; j += i) vis[j] = 1; vis[1] = 1; //注意}int main(){ int n, i; // 注意14 = 7 + 7 init(); while( ~scanf("%d", &n)) Read More

posted @ 2012-12-04 21:18 To be an ACMan Views(144) Comments(0) Diggs(0)

HDU 1319 Prime Cuts 素数水题

View Code #include<cstdio>#include<cstring>#include<cmath>int p[1004], tot;bool vis[1104];void init() // 素数打表，注意本题1也是素数{ int i, j; for(i = 2; i * i <= 1000; i++) for(j = i*i; j <= 1000; j += i) vis[j] = 1; for(i = 1; i <= 1000; i++) if(!vis[i])p[tot++] = i; }int mai... Read More

posted @ 2012-12-04 21:09 To be an ACMan Views(249) Comments(0) Diggs(0)

HDU 1576 A/B 扩展欧几里德算法模线性方程入门题

很详细的资料：http://blog.csdn.net/lulipeng_cpp/article/details/7612490补充以下结论，自己推的，解释了以上博客里的疑惑。方程ax+by=gcd(a,b)，即模线性方程ax≡d(mod b) ，令d =gcd(a,b)。假设模线性方程的解为 x0， y0。结论1：则有 max( abs(x0),abs(y0) )< max( abs(a), abs(b) );结论2：若d = a，则 x = 1，y = 0；若 d = b，则 x = 0，y = 1；结论3：则方程的所有解为 x = x0 + b/d*i, y = y0 - a Read More

posted @ 2012-12-04 19:55 To be an ACMan Views(436) Comments(0) Diggs(0)

2012年12月3日

HDU 2964 Prime Bases 数论水

View Code #include<cstdio>#include<cstring>int p[15] = {1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29};int a[15];int main(){ int i, j, k, n; while( ~scanf("%d", &n) && n) { printf("%d = ", n); int s = 1; for(i = 0; i < 13; i++) { s *= p[i]; if(... Read More

posted @ 2012-12-03 21:40 To be an ACMan Views(214) Comments(0) Diggs(0)

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