Problem 3 \u6570\u4f4d\u548c\u4e58\u79ef(digit.cpp\/c\/pas)<\/p>\n
\u3010\u9898\u76ee\u63cf\u8ff0\u3011<\/p>\n
\u4e00\u4e2a\u6570\u5b57\u7684\u6570\u4f4d\u548c\u4e58\u79ef\u4e3a\u5176\u5404\u4f4d\u6570\u5b57\u7684\u4e58\u79ef\u3002\u6c42\u6240\u6709\u7684N\u4f4d\u6570\u4e2d\u6709\u591a\u5c11\u4e2a\u6570\u7684\u6570\u4f4d\u548c\u4e58\u79ef\u6070\u597d\u4e3aK\u3002\u8bf7\u6ce8\u610f\uff0c\u8fd9\u91cc\u7684N\u4f4d\u6570\u662f\u53ef\u4ee5\u6709\u524d\u5bfc\u96f6\u7684\u3002\u6bd4\u598201\uff0c02\u89c6\u4e3a\u4e8c\u4f4d\u6570\uff0c\u4f46\u662f\u4ed6\u4eec\u7684\u6570\u4f4d\u548c\u4e58\u79ef\u90fd\u662f0\u3002<\/p>\n
\u3010\u8f93\u5165\u683c\u5f0f\u3011<\/p>\n
\u4e00\u884c\u4e24\u4e2a\u6574\u6570N\uff0cK<\/p>\n
\u3010\u8f93\u51fa\u683c\u5f0f\u3011<\/p>\n
\u4e00\u4e2a\u884c\u4e00\u4e2a\u6574\u6570\u8868\u793a\u7ed3\u679c\u3002<\/p>\n
\u3010\u6837\u4f8b\u8f93\u5165\u3011<\/p>\n
2 3<\/p>\n
\u3010\u6837\u4f8b\u8f93\u51fa\u3011<\/p>\n
2<\/p>\n
\u3010\u6837\u4f8b\u8f93\u51652\u3011<\/p>\n
2 0<\/p>\n
\u3010\u6837\u4f8b\u8f93\u51fa2\u3011<\/p>\n
19<\/p>\n
\u3010\u6570\u636e\u8303\u56f4\u3011<\/p>\n
\u5bf9\u4e8e20%\uff1aN <= 6\u3002<\/p>\n
\u5bf9\u4e8e50%\uff1aN<=16<\/p>\n
\u5b58\u5728\u53e6\u591630%\uff1aK=0\u3002<\/p>\n
\u5bf9\u4e8e100%\uff1aN <= 50\uff0c0 <= K <= 10^9\u3002<\/p>\n
\u6cd51\uff1a<\/p>\n
k=0\u65f6\uff0cans=10^n-9^n<\/p>\n
\u5426\u5219\u5c31\u7528\u8bb0\u5fc6\u5316\u641c\u7d22\u586b1..9\u7684\u4e2a\u6570<\/p>\n
\n
} } }; \u6cd52\uff1a<\/p>\n f[i][j][k][l][m]\u8868\u793a\u586b\u5230\u7b2ci\u4e2a\u6570,2,3,5,7\u5206\u522b\u7528j,k,l,m\u4e2a\u7684\u65b9\u6848\u6570<\/p>\n \u8003\u8651\u540e\u9762\u586b1..9\u7684\u60c5\u51b5(\u663e\u7136\u4e0d\u53ef\u80fd\u586b0)<\/p>\n Bug\uff1an=50,C(n,m)\u6700\u5927\u4e3aC(50,25),\u53ef\u7528long long.<\/p>\n ----<\/p>\n ----<\/p>\n \u6cd53\uff1a<\/p>\n \u5bb9\u6613\u53d1\u73b0<\/p>\n k=2^a*3^b*5^c*7^d\u65f6\u624d\u6709\u89e3<\/p>\n \u800c\u4e145,7\u53d6\u4e86\u5f53\u4e14\u53ea\u5f53\u6570\u4f4d\u4e3a5,7\u7684\u60c5\u51b5.<\/p>\n 2-2,4,6,8<\/p>\n 3-3,6,9<\/p>\n 5-5<\/p>\n 7-7<\/p>\n \u6545\u53ef\u53ea\u8003\u86512,3,\u4e0d\u8003\u86515,7<\/p>\n f[i][j]k]\u8868\u793ai\u4f4d\uff0c\u53d6\u4e86j\u4e2a2\u548ck\u4e2a3\u540e\u7684\u65b9\u6848\u6570\uff0c<\/p>\n \u56e0\u4e3a\u53ef\u4ee5\u75281\u586b\u5145\uff0c<\/p>\n \u6240\u4ee5\u7b54\u6848=\u2211f[p][j][k]*C(p,j+k)*C(n,a[5])*C(n-a[5],a[7]) (p<=n-a[5]-a[7])<\/p>\n <\/hasget><\/cctype><\/cmath><\/cstdlib><\/iostream><\/functional><\/algorithm><\/cstring><\/cstdio><\/code><\/p>\n","protected":false},"excerpt":{"rendered":" Problem 3 \u6570\u4f4d\u548c\u4e58\u79ef(digit.cpp\/c\/pas) \u3010\u9898\u76ee\u63cf\u8ff0\u3011 \u4e00\u4e2a\u6570\u5b57\u7684\u6570\u4f4d\u548c\u4e58\u79ef\u4e3a\u5176\u5404\u4f4d\u6570 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_coblocks_attr":"","_coblocks_dimensions":"","_coblocks_responsive_height":"","_coblocks_accordion_ie_support":"","footnotes":""},"categories":[12],"tags":[110,126],"class_list":["post-42","post","type-post","status-publish","format-standard","hentry","category-algorithm","tag-110","tag-126"],"yoast_head":"\n#include
\n#include
\n#include
\n#include
\n#include
\n#include
\n#include
\n#include
\nusing namespace std;
\n#define MAXN (50+10)
\n#define F (10000)
\nint n,k;
\nstruct highn
\n{
\n int len,a[900];
\n highn():len(1){memset(a,0,sizeof(a));}
\n highn(int x)
\n {
\n memset(a,0,sizeof(a));
\n int i=0;
\n while (x)
\n {
\n a[++i]=x%F;
\n x\/=F;
\n }
\n len=i;
\n }
\n void print()
\n {
\n printf(\"%d\",a[len]);
\n for (int i=len-1;i;i--)
\n {
\n \/\/ if (a[i]<10000) printf(\"0\");
\n if (a[i]<1000) printf(\"0\");
\n if (a[i]<100) printf(\"0\");
\n if (a[i]<10) printf(\"0\");
\n printf(\"%d\",a[i]);
\n }<\/p>\n
\n friend highn operator+(highn& a,highn& b)
\n {
\n highn c;
\n int maxlen=max(a.len,b.len);
\n for (int i=1;i< =maxlen;i++)\n {\n c.a[i]=c.a[i]+a.a[i]+b.a[i];\n if (c.a[i]>F)
\n {
\n c.a[i+1]+=c.a[i]\/F;
\n c.a[i]%=F;
\n }
\n }
\n c.len=maxlen+1;
\n while (!c.a[c.len]&&c.len>1) c.len--;
\n return c;<\/p>\n
\n friend highn operator-(highn& a,highn& b)
\n {
\n highn c;
\n int maxlen=a.len;
\n for (int i=1;i< =maxlen;i++)\n {\n c.a[i]+=a.a[i]-b.a[i];\n if (c.a[i]<0)\n {\n c.a[i+1]--;\n c.a[i]+=F;\n }\n }\n c.len=maxlen;\n while (!c.a[c.len]&&c.len>1) c.len--;
\n return c;
\n }
\n friend highn operator*(highn& a,highn& b)
\n {
\n highn c;
\n for (int i=1;i< =a.len;i++)\n {\n for (int j=1;j<=b.len;j++)\n {\n c.a[i+j-1]+=a.a[i]*b.a[j];\n if (c.a[i+j-1]>F)
\n {
\n c.a[i+j]+=c.a[i+j-1]\/F;
\n c.a[i+j-1]%=F;
\n }
\n }
\n }
\n c.len=a.len+b.len+1;
\n while (!c.a[c.len]&&c.len>1) c.len--;
\n return c;
\n }<\/p>\n
\nhighn pow2(highn a,int b)
\n{
\n highn c;
\n if (!b) return 1;
\n if (b%2)
\n {
\n c=pow2(a,b\/2);
\n c=c*c;
\n c=c*a;
\n }
\n else
\n {
\n c=pow2(a,b\/2);
\n c=c*c;
\n }
\n return c;
\n}
\nint a[11],tot,b[11];
\nhighn ans,C[51][51];
\nvoid dfs(int deep,highn rel,int hasget)
\n{
\n if (n