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銆銆璁綿1 = gcd(a, b), d2 = gcd(b, a % b), a > b涓攁 = qb + r
銆銆鍙渶璇乨1 | d2騫朵笖d2 | d1銆?br>銆銆鍥犱負(fù)d2 | b, d2 | r, 鍥犳d2 | (b + qr) = a錛屾牴鎹甦2 | b涓攄2 | a鏈塪2 | gcd(a, b)錛屽嵆d2 | d1銆?br>銆銆鍥犱負(fù)d1 | a, d1 | b, 鍥犳d1 | (a - qb) = r錛屾牴鎹甦1 | b涓攄1 | r鏈塪1 | gcd(b, r)錛屽嵆d1 | d2銆?br>銆銆鏍規(guī)嵁Euclid綆楁硶鎵ц榪囩▼錛実cd(a, b) = gcd(b, r) = gcd(r, b % r) = ... = gcd(rn, rn-1 % rn)錛屽鏋渞n-1 % rn = 0錛屾牴鎹甮cd(a, 0) = a錛屾湁gcd(a, b) = rn錛岃瘉姣曘?br>
銆銆璨屼技鍋忓簭鍏崇郴涓婅瘉a = b寰堝閮芥槸鍒╃敤鍙嶅縐版э紝姣斿a >= b騫朵笖b >= a鍒檃 = b錛屾垨鑰卆 | b騫朵笖b | a鍒檃 = b錛屼竴涓緢甯哥敤涓斿己澶х殑鏂規(guī)硶銆?br>
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銆愰鐩垎鏋愩?br>銆銆榪欐槸涓閬揃T棰橈紝涓棿璁劇疆浜?jiǎn)璁稿trick銆傚紑濮嬪棰樻剰娌℃湁瀹屽叏鐞嗚В錛岄敊浜?jiǎn)寰堝娆″Q屽悗鏉ユ壘鏉ヤ簡(jiǎn)鏁版嵁錛屾墠鍙戠幇浜?jiǎn)闂銆?br>銆銆棰樼洰涓弿榪扮殑鑾瘋儨絳栫暐鏄細(xì)"A player wins if after his move the size of some heap is a prime number."榪欏彞璇濅箥涓鐪嬩互涓烘槸鍙栧畬鐭沖瓙鍚庡墿涓嬬殑鐭沖瓙涓暟鏄礌鏁扮殑鏃跺欏氨鑾瘋儨錛屽叾瀹炶繕闅愯棌鐫鍙︿竴縐嶅彲鑳斤細(xì)濡傛灉澶氬爢鐭沖瓙涓暟鏄礌鏁幫紝褰撳墠鐜╁鏃犺鎬庢牱鍙栭兘鑳借幏鑳滐紝鍥犱負(fù)鍦ㄤ粬鍙栧畬涔嬪悗錛屽叾浠栧爢鐨勭煶瀛愪釜鏁版槸绱犳暟錛屼篃婊¤凍鑾瘋儨鏉′歡銆?br>銆銆鎺ヤ笅鏉ヨ冭檻涓鑸儏鍐點(diǎn)傝繖涓鐩槸闄愬埗涓棿鐘舵佺殑Nim娓告垙錛屼篃灝辨槸璇達(dá)紝瀵逛簬涓鍫嗕釜鏁頒負(fù)n鐨勭煶瀛愯岃█錛屽畠鐨凷G鍊煎彇鍐充簬灝忎簬n鐨勬渶澶х礌鏁般傛敞鎰忚繖閲岄璁懼張鏈変簡(jiǎn)涓涓皬trick錛岄鐩鏄庝簡(jiǎn)闇瑕佸彇1鍒発涓紝濡傛灉褰撳墠鐭沖瓙涓暟鏈韓鏄礌鏁幫紝褰撶劧鏄病鐢ㄧ殑錛屽洜姝ゆ槸灝忎簬n鐨勬渶澶х礌鏁般傝灝忎簬n鐨勬渶澶х礌鏁版槸p(棰樼洰涓鏄庝簡(jiǎn)鐭沖瓙涓暟澶т簬2錛屽洜姝涓瀹氬瓨鍦?錛岄偅涔堝彲浠ュ湪k姝ヤ互鍐呭埌杈緋鐨勪竴瀹氭槸蹇呰儨鎬侊紝鑰屼笖鏄洿鎺ヨ幏鑳滐紝闇瑕佸湪杈撳叆鐨勬椂鍊欑壒鍒?榪欎竴鐐歸渶瑕佹敞鎰忥紝鍦ㄨВ鍐抽檺鍒朵腑闂寸姸鎬佺殑Nim娓告垙鏃朵竴鑸兘闇瑕佺壒鍒?銆傜劧鍚庡氨鏄痯 + k + 1榪欎釜鐘舵侊紝鍥犱負(fù)瀹冨彲杈劇殑鐘舵佸叏鏄繀鑳滄侊紝鍥犳瀹冩槸蹇呰觸鎬侊紝SG鍊間負(fù)0銆傜幇鍦ㄨ冭檻澶т簬p + k + 1鐨勭姸鎬侊紝闂鍑烘潵浜?jiǎn)銆備互p + k + 2榪欎釜鐘舵佷負(fù)渚嬶紝鍥犱負(fù)瀹冨彲浠ュ埌杈緋 + 2 ... p + k + 1榪欎簺鐘舵侊紝鍥犱負(fù)p + 2 ... p + k閮芥槸鐩存帴鑾瘋儨鐘舵侊紝濡備綍鍒ゅ畾浠栦滑鐨凷G鍊煎憿錛熷鏋滃亣璁懼畠浠殑SG鍊兼槸1錛岄偅涔坧 + k + 2榪欎釜鐘舵佺殑SG鍊煎簲璇ユ槸2銆備絾鏄濊冧竴涓婼G鍊肩殑瀹氫箟錛屽畠鏄畾涔夊湪涓涓狣AG涓婄殑錛屾墍鏈夌殑鐘舵佹渶鍚庨兘鏄彲浠ュ湪鏈夐檺姝ュ唴杞Щ鍒扮粓姝㈡?蹇呰觸鎬?銆備絾鏄痯 + 1 ... p + k榪欎簺鐘舵侀兘杞Щ鍒頒簡(jiǎn)p榪欎釜鐘舵佷笂錛屾垜浠偗瀹氫笉鑳借瀹歱鐘舵佹槸緇堟鎬侊紝鍥犳浠呬粎鍑焢 + 1 ... p + k榪欎簺鐘舵佹槸蹇呰儨鎬佸氨綆鍗曠殑鎶婂畠浠殑SG鍊艱涓?鏄笉鎭板綋鐨勶紱榪欎簺闄愬埗鐘舵佸拰浠ュ墠鐨勯鐩繕?shù)笉鍚屽Q岃繖浜涢檺鍒剁姸鎬侀兘涓嶈兘杞Щ鍒扮粓姝㈡佷笂錛屼絾鏄敱浜庨鐩殑瑕佹眰錛屽畠浠張閮芥槸蹇呰儨鎬侊紝鍥犳鎶婂畠浠殑SG鍊艱涓烘棤絀峰ぇ鏇村悎閫備簺銆?br>銆銆浠旂粏鎬濊冧竴涓嬪甫闄愬埗鐘舵佺殑SG娓告垙錛屽彲浠ュ彂鐜幫紝瀹冧滑鍜屼竴鑸殑SG娓告垙鐨勫尯鍒湪浜庯紝鍦ㄥ垎鏋愪竴鑸殑SG娓告垙鐨勬椂鍊欙紝瀵逛簬涓涓姸鎬佸浘鑰岃█錛岃漿縐誨埌緇堟鎬佺殑鏃跺欏茍涓嶆剰鍛崇潃娓告垙緇撴潫錛屽洜涓虹帺瀹跺彲浠ラ氳繃璧板叾浠栫殑鐘舵佸浘鏉ヤ繚璇佹槸鍚﹁揪鍒板繀鑳滄侊紱浣嗘槸甯﹂檺鍒剁姸鎬佺殑SG娓告垙錛岄檺鍒剁殑鐘舵佸弻鏂歸兘鏄笉鏁㈣蛋鐨勶紝鍥犱負(fù)涓鏃︿竴鏂硅蛋鍏ラ檺鍒剁姸鎬侊紝鍙︿竴鏂圭珛鍒昏幏鑳滐紝娓告垙灝辯粓姝簡(jiǎn)銆傚彲浠ヨ涓猴紝鍙璧板叆闄愬埗鎬佸氨鐩稿綋浜庤杈擄紝瀵逛簬鍙屾柟鐜╁鑰岃█鑲畾閮戒笉浼?xì)杩欐牱鍋氬Q屽洜姝よ繖浜涢檺鍒剁姸鎬佸氨鎴愪簡(jiǎn)“姝葷姸鎬?#8221;錛屽畬鍏ㄥ彲浠ュ拷瑙嗚繖浜涚姸鎬侊紙涔熷氨鏄涓嶅瓨鍦ㄥ埌榪欎簺鐘舵佺殑杞Щ錛夈傞氳繃涓婅堪鍒嗘瀽錛屾垜浠彲浠ヨ瀹歱 + k + 2榪欎釜鐘舵佺殑SG鍊煎簲璇ユ槸1鑰屼笉鏄?銆?br>銆銆鐜板湪榪欎釜闂鐨勫仛娉曞氨姣旇緝鏄庢湕浜?jiǎn)锛屽浜庢瘡鍫嗙煕_瓙錛屽幓鎺夐檺鍒舵佺殑璁ㄨ鍚庯紝灝卞彉鎴愪簡(jiǎn)鍦ㄩ泦鍚坽1...k}涓夋嫨鍏冪礌鐨勪竴涓猄ubtraction Game錛屽畠鐨凷G鍊兼槸妯?k + 1)寰幆鐨勩?br>銆銆鐒跺悗灝辨槸姹傛渶榪戠礌鏁扮殑闂?shù)簡(jiǎn)锛寴q欎釜娌℃湁濂藉姙娉曪紝鍙兘鏆村姏鏋氫婦銆傚浜庝竴涓暟n錛屽湪sqrt(n)鍒皀涔嬮棿瀛樺湪绱犳暟錛堟劅瑙夊簲璇ユ槸錛屼絾鏄笉浼?xì)璇佸Q夛紝鍥犳鏈澶氭灇涓緎qrt(n)嬈″氨鑳芥壘鍒拌В銆備絾鏄瘡嬈℃灇涓懼垽绱犵殑澶嶆潅搴﹁繕鏄痵qrt(n)錛屾誨鏉傚害榪樻槸姣旇緝楂橈紝鎴戝湪鏈湴璺戞暟鎹窇浜?縐掞紝浜や笂鍘昏秴鏃朵簡(jiǎn)錛圫POJ鐨勬椂闄?0縐掞紝榪欏眳鐒朵篃瓚呮椂錛屾暟鎹鍙樻侊級(jí)銆傛兂浜?jiǎn)寰堜箙涔熸病鏈変粈涔堟柊鐨勭畻娉曪紝鏃犲鍙兘鍦ㄥ垽绱犱笂涓嬬偣鍔熷か錛岀洿鎺ユ妸Miller-Rabin鎼炲嚭鏉ヤ簡(jiǎn)錛岄熷害鎻愰珮鍒頒簡(jiǎn)2縐掞紝浠嶇劧瓚呮椂銆傚悗鏉ュ張鍔犱簡(jiǎn)涓浜涘父鏁頒笂鐨勪紭鍖栵紝緇堜簬榪囦簡(jiǎn)銆?br>
銆愭葷粨銆?br>銆銆榪欎釜棰樼洰鍋氫簡(jiǎn)寰堥暱鏃墮棿錛屼竴鏂歸潰鏄棰樹(shù)笉娓咃紝閿欎簡(jiǎn)鍑犳錛涘彟涓鏂歸潰鏄浜嶴G鐨勭悊璁虹悊瑙ｇ殑涓嶅閫忓交錛屾兂浜?jiǎn)寰堜箙缁堜簬鎯虫槑鐧戒簡(jiǎn)锛涘啀鏈墳鏄紭鍖柦帡娉曚篃鑺变簡(jiǎn)寰堥暱鏃堕棿銆備笉榪囬氳繃榪欎釜棰樼洰瀵逛簬SG鐞嗚鐨勭悊瑙ｅ張榪涗簡(jiǎn)涓姝ワ紝鎰熻涓嶉敊錛屽懙鍛點(diǎn)?
娉細(xì)鏈枃浣滀簬2009騫?鏈?鏃?9鐐?7鍒?br>
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銆愰鐩垎鏋愩?br>銆銆榪欎釜棰樼洰涔嶄竴鐪嬫尯綆鍗曠殑錛屼絾鏄兂鍋氬榪樻槸瑕佷粩緇嗙殑鎬濊冧笅鐨勩傝ˉ闆嗚漿鍖栫殑鎬濇兂錛屾眰鍑烘墍鏈夊叡綰跨殑涓夊厓緇勶紝鐒跺悗鐢ㄦ繪暟鍑忔帀灝辨槸絳旀浜?jiǎn)锛屽叧閿疄鏄浣曟眰鍏本U夸笁鍏冪粍銆倄鍧愭爣鐩稿悓鍜寉鍧愭爣鐩稿悓鐨勬瘮杈冨ソ璁＄畻錛屽湪涓鏉℃枩綰跨殑灝變笉濂界畻浜?jiǎn)锛岀斨M釜鍥懼彂鐜幫紝鍗充嬌鏂滅巼鐩稿悓鐨勭嚎錛岀粡榪囩殑鏍肩偣鏁板彲鑳藉悇涓嶇浉鍚屻傛濊礬褰撶劧榪樻槸鏋氫婦y / x錛堜笉鍚岀殑y / x紜畾浜?jiǎn)涓嶅悓鐨勭煩迮炲尯鍩熷Q夛紝涔嬪悗濡備綍鏈夋晥鐨勮綆楋紝鎴戦噰鐢ㄧ殑鏂規(guī)硶鍙兘鏈変簺楹葷儲(chǔ)錛屾湁鐐圭被瀹規(guī)枼鐨勬柟娉曘備互鏂滅巼涓篴 / b涓轟緥錛?a, b) = g錛岄偅涔坢 * n鐨勫尯鍩熷唴涓瀹氭湁(m - a + 1) * (n - b + 1)涓偅涔堝ぇ鐨勭煩褰紝榪欐牱鐨勭煩褰㈢粡榪囩殑鏍肩偣鏁版槸(g + 1)錛涚劧鍚庡洜涓哄悓絳夋枩鐜囧皬涓鐐圭殑鐭╁艦(a - a / g, b - b / g)涔熸槸瀛樺湪鐨勶紝涓暟鍚屾牱鍙互緇熻鍑烘潵錛屼絾鏄湁浜涘ぇ鐭╁艦鍖呮嫭浜?jiǎn)锛岃鍘绘帀锛涘洜姝閲囩敤杩櫩U嶆濇兂錛屽厛姹傚嚭澶х煩褰㈢殑涓暟錛岀劧鍚庝緷嬈″線(xiàn)涓嬪噺錛屽氨鍙互閬垮厤閲嶅璁℃暟浜?jiǎn)銆傝櫧鐒惰繖鏍峰鏉傚害鏈夌偣楂橈紝涓嶈繃鏋侀檺鏁版嵁榪樻槸姣旇緝蹇殑璺戝嚭鏉ヤ簡(jiǎn)銆?br>銆銆璇寸殑鍙兘涓嶅お娓呮錛屽叿浣撲唬鐮佸涓嬶細(xì)
1 #include <cstdio>
2 #include <iostream>
3 using namespace std;
4 const int N = 1024;
5
6 bool tag[N][N];
7 long long calc(long long n)
8 {
9 if (n <= 2)
10 return 0;
11 return n * (n - 1) * (n - 2) / 6;
12 }
13 int gcd(int a, int b)
14 {
15 return b == 0 ? a : gcd(b, a % b);
16 }
17
18 int main()
19 {
20 int m, n, g, a, b, timer = 1, cnt[N], ta, tb;
21 long long ans;
22
23 while (scanf("%d %d", &m, &n) == 2)
24 {
25 memset(tag, 0, sizeof(tag));
26 if (m == 0 && n == 0)
27 break;
28 ans = calc((m + 1) * (n + 1)) - calc(m + 1) * (n + 1) - calc(n + 1) * (m + 1);
29 for (int i = m; i >= 1; i--)
30 for (int j = n; j >= 1; j--)
31 {
32 g = gcd(i, j);
33 a = i / g, b = j / g;
34 if (tag[a][b]) continue;
35 memset(cnt, 0, sizeof(cnt));
36 tag[a][b] = 1;
37 a = i, b = j;
38 ta = i / g, tb = j / g;
39 for (int k = g; k >= 2; k--)
40 {
41 cnt[k] += (m - a + 1) * (n - b + 1);
42 ans -= calc(k + 1) * cnt[k] * 2;
43 for (int t = 1; t <= k - 2; t++)
44 cnt[k-t] -= (t + 1) * cnt[k];
45 a -= ta, b -= tb;
46 }
47 }
48 cout << "Case " << timer++ << ": " << ans << endl;
49 }
50
51 return 0;
52 }
53
2 #include <iostream>
3 using namespace std;
4 const int N = 1024;
5
6 bool tag[N][N];
7 long long calc(long long n)
8 {
9 if (n <= 2)
10 return 0;
11 return n * (n - 1) * (n - 2) / 6;
12 }
13 int gcd(int a, int b)
14 {
15 return b == 0 ? a : gcd(b, a % b);
16 }
17
18 int main()
19 {
20 int m, n, g, a, b, timer = 1, cnt[N], ta, tb;
21 long long ans;
22
23 while (scanf("%d %d", &m, &n) == 2)
24 {
25 memset(tag, 0, sizeof(tag));
26 if (m == 0 && n == 0)
27 break;
28 ans = calc((m + 1) * (n + 1)) - calc(m + 1) * (n + 1) - calc(n + 1) * (m + 1);
29 for (int i = m; i >= 1; i--)
30 for (int j = n; j >= 1; j--)
31 {
32 g = gcd(i, j);
33 a = i / g, b = j / g;
34 if (tag[a][b]) continue;
35 memset(cnt, 0, sizeof(cnt));
36 tag[a][b] = 1;
37 a = i, b = j;
38 ta = i / g, tb = j / g;
39 for (int k = g; k >= 2; k--)
40 {
41 cnt[k] += (m - a + 1) * (n - b + 1);
42 ans -= calc(k + 1) * cnt[k] * 2;
43 for (int t = 1; t <= k - 2; t++)
44 cnt[k-t] -= (t + 1) * cnt[k];
45 a -= ta, b -= tb;
46 }
47 }
48 cout << "Case " << timer++ << ": " << ans << endl;
49 }
50
51 return 0;
52 }
53
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#include <cstdio>#include <cmath>#include <algorithm>using namespace std;const int MOD = 11, N = 100001;int d[N], m[N];int power_mod(int a, int b)
{
int ret = 1, f = a; while (b)
{ if (b & 1) ret = ret * f % MOD; f = f * f % MOD; b >>= 1;
} return ret;
}void init(){ int tmp; for (int i = 2; i < N; i++) { tmp = (int)(ceil(i * log(i) / log(10.0) + 1e-2) + 1e-1); d[i] = tmp; m[i] = power_mod(i % MOD, i); }}int main(){ char str[N*5]; int T, p, len, tmp; init(); scanf("%d", &T); while (T--) { scanf("%s", str); len = strlen(str); if (len == 1 && str[0] == '1') { puts("1"); continue; } p = lower_bound(d, d + N, len) - d; tmp = 0; for (int i = 0; i < len; i++) { tmp = tmp * 10 + (str[i] - '0'); tmp = tmp % MOD; } printf("%d\n", tmp == m[p] ? p : -1); } return 0;}]]>
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銆銆浠 = (z - y) / 2錛寁 = (z + y) / 2銆傚洜涓簓銆亃浜掕川錛岄偅涔坲銆乿浜掕川銆傚鏋滀笉浜掕川錛屾湁v = ku錛屽彉鎹㈠悗寰楀埌錛歾 / y = (k + 1) / (k - 1)銆傚洜涓簔鍜寉浜掕川錛岃繖鏍瘋繖涓瓑寮忕殑瑙ｅ彧鑳芥槸z = k + 1涓攜 = k - 1銆傝繖鏍風(fēng)殑璇漸 = 1銆傚彲浠ョ悊瑙ｄ負(fù)榪欎篃鏄簰璐ㄧ殑鐗規(guī)畩鎯呭喌(gcd = 1)銆傝繖鏍?x / 2) ^ 2 = u * v錛寀鍜寁娌℃湁鍏叡鐨勮川鍥犳暟錛屽洜姝ゅ繀鐒秛鍜寁閮芥槸瀹屽叏騫蟲(chóng)柟鏁般傛帴涓嬫潵鐨勪簨鎯呭氨姣旇緝綆鍗曚簡(jiǎn)錛屼護(hù)u = m ^ 2錛寁 = n ^ 2錛岀劧鍚庡埄鐢╩鍜宯琛ㄧずx銆亂銆亃鏈夛細(xì)y = n ^ 2 - m ^ 2錛寈 = 2 * m * n錛寊 = n ^ 2 + m ^ 2銆傝繖鏍烽鐩腑鏁版嵁鑼冨洿1000000錛岃屾灇涓緈鍜宯鐨勮寖鍥存渶澶?000錛岃繖涓鏉傚害灝卞彲浠ユ帴鍙椾簡(jiǎn)銆傛灇涓劇殑鏃跺欎笉鐢ㄦ媴蹇?jī)鍑虹幇閲嶅鐨勪簰璐ㄧ殑x涓巠錛屽洜涓鴻繖閲寈涓瀹氭槸鍋舵暟錛屽鏋渪鍜寉宸茬粡浜掕川錛岄偅涔坹涓瀹氭槸濂囨暟錛岃В鏄敮涓鐨勩?br />銆銆寰堝洤鐨勬槸榪欎釜宸у鐨勬柟娉曞眳鐒跺叕鍏冨墠250騫村氨鏈夊ぇ鐗涙兂鍒頒簡(jiǎn)錛堜涪鐣浘錛夛紝瀹炲湪orz錛?
]]>
闄凱KU 1811浠ｇ爜錛?/p>
PKU 18111 #include <cstdio>
2 #include <stdlib.h>
3 #include <time.h>
4 const int MAX = 8, N = 1000010;
5
6 long long ans, p[N/5];
7 int tag[N];
8
9 long long abs(long long t)
10 {
11 return t < 0 ? -t : t;
12 }
13 void prime()
14 {
15 int cnt = 0;
16
17 for (int i = 2; i < N; i++)
18 {
19 if (!tag[i]) p[cnt++] = i;
20 for (int j = 0; j < cnt && p[j] * i < N; j++)
21 {
22 tag[i*p[j]] = p[j];
23 if (i % p[j] == 0)
24 break;
25 }
26 }
27 }
28 long long gcd(long long a, long long b)
29 {
30 return !b ? a : gcd(b, a % b);
31 }
32 long long MultiMod(long long a, long long b, long long k)
33 {
34 long long ret = 0, tmp = a % k;
35
36 while (b)
37 {
38 if (b & 1)
39 ret = (ret + tmp) % k;
40 tmp = (tmp << 1) % k;
41 b >>= 1;
42 }
43
44 return ret;
45 }
46 long long PowerMod(long long a, long long b, long long k)
47 {
48 long long ret = 1, f = a % k;
49
50 while (b)
51 {
52 if (b & 1)
53 ret = MultiMod(ret, f, k);
54 f = MultiMod(f, f, k);
55 b >>= 1;
56 }
57 return ret;
58 }
59 int TwiceDetect(long long a, long long b, long long k)
60 {
61 int t = 0;
62 long long x, y;
63
64 while ((b & 1) == 0)
65 {
66 b >>= 1;
67 t++;
68 }
69 y = x = PowerMod(a, b, k);
70 while (t--)
71 {
72 y = MultiMod(x, x, k);
73 if (y == 1 && x != 1 && x != k - 1)
74 return 0;
75 x = y;
76 }
77 return y != 1 ? 0 : 1;
78 }
79 bool MillerRabin(long long n)
80 {
81 int i;
82 long long tmp;
83
84 srand(time(0));
85 for (i = 0; i < MAX; i++)
86 {
87 tmp = rand() % (n - 1) + 1;
88 if (TwiceDetect(tmp, n - 1, n) != 1)
89 break;
90 }
91 return (i == MAX);
92 }
93 long long pollard_rho(long long n)
94 {
95 srand(time(0));
96 int i = 1, k = 2;
97 long long d, x = rand() % (n - 1) + 1, y = x;
98 while (true)
99 {
100 i++;
101 x = (MultiMod(x, x, n) - 1) % n;
102 d = abs(gcd(y - x, n));
103 if (d != 1 && d != n)
104 return d;
105 if (i == k)
106 {
107 y = x;
108 k <<= 1;
109 }
110 }
111 }
112 void factorization(long long n)
113 {
114 long long tmp;
115
116 if (n < N)
117 {
118 ans <?= !tag[n] ? n : tag[n];
119 return;
120 }
121 if (MillerRabin(n))
122 {
123 ans <?= n;
124 return;
125 }
126 tmp = pollard_rho(n);
127 factorization(tmp);
128 factorization(n / tmp);
129 }
130
131 int main()
132 {
133 int T;
134 long long n;
135
136 scanf("%d", &T);
137 prime();
138 while (T--)
139 {
140 scanf("%I64d", &n);
141 if (n < N)
142 {
143 if (tag[n] == 0)
144 puts("Prime");
145 else
146 printf("%d\n", tag[n]);
147 continue;
148 }
149 if (MillerRabin(n))
150 puts("Prime");
151 else
152 {
153 ans = 0xfffffffffffffffLL;
154 factorization(n);
155 printf("%I64d\n", ans);
156 }
157 }
158
159 return 0;
160 }
161
]]>
1 #include <cstdio>
2 const int M = 14;
3 const long long max = 1000000000000000LL;
4
5 int p[M] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41}, k;
6 long long ans;
7 void solve(long long v, int factor, int pos)
8 {
9 if (factor >= k)
10 {
11 if (factor == k) ans <?= v;
12 return;
13 }
14 if (k % factor) return;
15 if (pos == M) return;
16 for (int i = 1; i <= 50; i++)
17 {
18 v *= p[pos];
19 if (v > max) break;
20 solve(v, factor * (i + 1), pos + 1);
21 }
22 }
23
24 int main()
25 {
26 while (scanf("%d", &k) == 1)
27 {
28 ans = max + 1;
29 solve(1, 1, 0);
30 if (ans > max) printf("-1\n");
31 else printf("%lld\n", ans);
32 }
33
34 return 0;
35 }
36
2 const int M = 14;
3 const long long max = 1000000000000000LL;
4
5 int p[M] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41}, k;
6 long long ans;
7 void solve(long long v, int factor, int pos)
8 {
9 if (factor >= k)
10 {
11 if (factor == k) ans <?= v;
12 return;
13 }
14 if (k % factor) return;
15 if (pos == M) return;
16 for (int i = 1; i <= 50; i++)
17 {
18 v *= p[pos];
19 if (v > max) break;
20 solve(v, factor * (i + 1), pos + 1);
21 }
22 }
23
24 int main()
25 {
26 while (scanf("%d", &k) == 1)
27 {
28 ans = max + 1;
29 solve(1, 1, 0);
30 if (ans > max) printf("-1\n");
31 else printf("%lld\n", ans);
32 }
33
34 return 0;
35 }
36
]]>
璁緁[i]琛ㄧずi鐨勭害鏁板拰錛宔[i]琛ㄧずi鐨勬渶灝忕礌鍥犲瓙涓暟錛宼[i]琛ㄧず(p1 ^ 0 + .. + p1 ^ a1)錛宲1鏄痶鐨勬渶灝忕礌鍥犲瓙錛宎1鏄痯1鐨勫箓嬈★紝榪欐牱瀵逛簬i * p[j]錛屽鏋減[j]涓嶆槸i鐨勫洜瀛愶紝閭ｄ箞鏍規(guī)嵁縐ф潯浠訛紝f[i*p[j]] = f[i] * (1 + p[j])錛宔[i] = 1錛宼[i] = 1 + p[j]錛涘鏋減[j]鏄痠鐨勫洜瀛愶紝閭ｄ箞鐩稿綋浜巘[i]澶氫簡(jiǎn)涓欏筽1 ^ (a1 + 1)錛岄鍏坋[i]++錛岀劧鍚巘mp = t[i]錛宼[i] += p[j] ^ e[i]錛宖[i*p[j]] = f[i] / tmp * t[i]銆傝繖鏍蜂篃灝卞仛鍒頒簡(jiǎn)O(1)鐨勬椂闂磋綆楀嚭浜?jiǎn)f[i*p[j]]鍚屾椂涔熻綆楀嚭浜?jiǎn)闄勫姞淇℃伅銆?br>
榪欑鏂規(guī)硶榪樺彲浠ョ戶(hù)緇帹騫匡紝渚嬪鍙互璁板綍i鐨勬渶灝忕礌鍥犲瓙錛岃繖鏍峰氨鍙互鍋氬埌O(log n)鏃墮棿鐨勭礌鍥犲瓙鍒嗚В銆?/p>
]]>
TJU 3027#include <cstdio>const int N = 8;void extended_gcd(int a, int b, int &x, int &y)
{ if (!b)
{ x = 1; y = 0; return; } extended_gcd(b, a % b, x, y); int tmp = x; x = y; y = tmp - a / b * y;}int modular_linear_equation_system(int b[N], int m[N], int n){ int M = 1, x, y, ret = 0; for (int i = 0; i < n; i++) M *= m[i]; for (int i = 0; i < n; i++) { extended_gcd(M / m[i], m[i], x, y); ret += M / m[i] * x * b[i]; } while (ret <= 0) ret += M; while (ret > M) ret -= M; return ret;}int main(){ int m[N], b[N], T, n, tmp, ans; const int num = 4, numc = 3; char str[N], hash[30] = " ABCDEFGHIJKLMNOPQRSTUVWXYZ "; scanf("%d", &T); while (T--) { scanf("%d", &n); for (int i = 0; i < num; i++) scanf("%d", &m[i]); for (int j = 0; j < n; j++) { scanf("%d", &tmp); for (int i = 1; i <= num; i++) { b[num-i] = tmp % 100; tmp /= 100; } ans = modular_linear_equation_system(b, m, num); for (int i = 1; i <= numc; i++) { str[numc-i] = hash[ans%100]; ans /= 100; } str[numc] = '\0'; if (j == n - 1) { int k = numc - 1; for (; k >= 0; k--) if (str[k] != ' ') break; str[k+1] = '\0'; } printf("%s", str); } putchar('\n'); } return 0;}]]>
#include <cstdio>#include <stdlib.h>const int MAX = 4, N = 1000000;long long PowerMod(long long a, long long b, long long k){ long long ret = 1, f = a; while (b) { if (b & 1) ret = ret * f % k; f = f * f % k; b >>= 1; } return ret;}bool MillerRabin(long long n){ int i; long long tmp; srand(100); for (i = 0; i < MAX; i++) { tmp = rand() % (n - 1) + 1; if (PowerMod(tmp, n - 1, n) != 1) break; } return (i == MAX);}int main(){ long long n, i, j; bool tag[N] = {1, 1, 0}; for (i = 2; i * i < N; i++) { if (tag[i]) continue; for (j = i; j * i < N; j++) tag[j*i] = 1; } while (scanf("%lld", &n) == 1) { if (n < N) printf("%s\n", tag[n] ? "NO" : "YES"); else printf("%s\n", ((n & 1) == 0) || !MillerRabin(n) ? "NO" : "YES"); } return 0;}]]>
#include <cstdio>
long long PowerMod(long long a, int b, int k)
{
long long tmp = a, ret = 1;
while (b)
{
if (b & 1)
ret = ret * tmp % k;
tmp = tmp * tmp % k;
b >>= 1;
}
return ret;
}
int main()
{
long long a;
int b, k;
while (scanf("%lld %d %d", &a, &b, &k) == 3)
printf("%lld\n", PowerMod(a, b, k));
return 0;
}
long long PowerMod(long long a, int b, int k)
{
long long tmp = a, ret = 1;
while (b)
{
if (b & 1)
ret = ret * tmp % k;
tmp = tmp * tmp % k;
b >>= 1;
}
return ret;
}
int main()
{
long long a;
int b, k;
while (scanf("%lld %d %d", &a, &b, &k) == 3)
printf("%lld\n", PowerMod(a, b, k));
return 0;
}
]]>