多多色-多人伦交性欧美在线观看-多人伦精品一区二区三区视频-多色视频-免费黄色视屏网站-免费黄色在线

國內最全IT社區平臺 聯系我們 | 收藏本站
阿里云優惠2
您當前位置:首頁 > 互聯網 > hdu 4472 Count (遞推)

hdu 4472 Count (遞推)

來源:程序員人生   發布時間:2014-11-09 08:27:45 閱讀次數:2263次

Count

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1756    Accepted Submission(s): 1133


Problem Description
Prof. Tigris is the head of an archaeological team who is currently in charge of an excavation in a site of ancient relics.
This site contains relics of a village where civilization once flourished. One night, examining a writing record, you find some text meaningful to you. It reads as follows.
“Our village is of glory and harmony. Our relationships are constructed in such a way that everyone except the village headman has exactly one direct boss and nobody will be the boss of himself, the boss of boss of himself, etc. Everyone expect the headman is considered as his boss’s subordinate. We call it relationship configuration. The village headman is at level 0, his subordinates are at level 1, and his subordinates’ subordinates are at level 2, etc. Our relationship configuration is harmonious because all people at same level have the same number of subordinates. Therefore our relationship is …”
The record ends here. Prof. Tigris now wonder how many different harmonious relationship configurations can exist. He only cares about the holistic shape of configuration, so two configurations are considered identical if and only if there’s a bijection of n people that transforms one configuration into another one.
Please see the illustrations below for explanation when n = 2 and n = 4.

The result might be very large, so you should take module operation with modules 109 +7 before print your answer.
 

Input
There are several test cases.
For each test case there is a single line containing only one integer n (1 ≤ n ≤ 1000).
Input is terminated by EOF.
 

Output
For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) and Y is the desired answer.
 

Sample Input
1 2 3 40 50 600 700
 

Sample Output
Case 1: 1 Case 2: 1 Case 3: 2 Case 4: 924 Case 5: 1998 Case 6: 315478277 Case 7: 825219749
 

對每一個合法圖形,記最底層個數為j,則再增加1層增加的個數必須是j的倍數。可以用dp[i][j]表示i個點最底層為j個時的個數,對數據范圍內的N遍歷得到答案。(屬于“我為人人”型的遞推關系)

dp[i][j]-->dp[i+j*k][j*k](k=1...N)


#include <iostream> #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> using namespace std; #define N 1010 #define LL __int64 const int mod=1000000007; int f[N][N],ans[N]; void inti() { int i,j,k; memset(f,0,sizeof(f)); f[1][1]=1; for(i=1;i<=1000;i++) { for(j=1;j<=1000;j++) { if(f[i][j]==0) //由當前合法狀態推得其他狀態 continue; for(k=1;k<=1000;k++) { int t1=j*k; int t2=t1+i; if(t2>1000) break; f[t2][t1]+=f[i][j]; if(f[t2][t1]>=mod) f[t2][t1]%=mod; } } } for(i=1;i<=1000;i++) { int tmp=0; for(j=1;j<=i;j++) tmp=(tmp+f[i][j])%mod; ans[i]=tmp; } } int main() { int n,cnt=1; inti(); while(scanf("%d",&n)!=⑴) { printf("Case %d: %d ",cnt++,ans[n]); } return 0; }


生活不易,碼農辛苦
如果您覺得本網站對您的學習有所幫助,可以手機掃描二維碼進行捐贈
程序員人生
------分隔線----------------------------
分享到:
------分隔線----------------------------
關閉
程序員人生
主站蜘蛛池模板: 亚洲视频免费播放 | 国产福利视频一区二区三区四区 | 伊人久久大香线蕉综合网站 | 日本护士和病人xxxxx | 波多野结衣中文字幕在线视频 | 日本一区二区高清不卡 | 波多野结衣视频在线看 | 中文字幕第12页 | 精品久久国产视频 | 黄色免费观看视频网站 | 精品久久久久国产免费 | 日本三级成人中文字幕乱码 | 国产一级毛片视频在线! | 天天天狠天天透天天制色 | аⅴ成人天堂中文在线 | 久久精品高清 | 日韩 亚洲 中文 图片 小说 | 伊人55| 五月天婷婷在线视频国产在线 | 成人免费在线视频网站 | 欧美最猛黑人xxxxx猛交 | 久久国| 成人毛片国产a | 亚洲国产日韩在线观频 | 在线观看国产免费高清不卡 | 欧美福利 | 国产免费一区二区在线看 | 欧美精品黄页免费高清在线 | 波多野结衣一区二区三区在线观看 | jizz欧美大全 | 欧美人与牲动交a欧美精品 欧美人与牲动交xxxx | 一级做a爰片性色毛片黄书 一级做a爰片性色毛片男 | 久久精品国产第一区二区 | 久久456| 日韩欧美亚洲一区精选 | 国产片在线观看 | 美国jizz| 日韩精品无码一区二区三区 | 天堂最新版在线www在线 | 国产v亚洲v天堂a无 国产v亚洲v天堂无码 | 久久国产精品-久久精品 |