MiYu鍘熷垱, 杞笘璇鋒敞鏄?: 杞澆鑷?nbsp;______________鐧界櫧銇眿 
棰樼洰鎻忚堪:
Cube
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 495 Accepted Submission(s): 226
Problem Description
Given an N*N*N cube A, whose elements are either 0 or 1. A[i, j, k] means the number in the i-th row , j-th column and k-th layer. Initially we have A[i, j, k] = 0 (1 <= i, j, k <= N).
We define two operations, 1: “Not” operation that we change the A[i, j, k]=!A[i, j, k]. that means we change A[i, j, k] from 0->1,or 1->0. (x1<=i<=x2,y1<=j<=y2,z1<=k<=z2).
0: “Query” operation we want to get the value of A[i, j, k].
We define two operations, 1: “Not” operation that we change the A[i, j, k]=!A[i, j, k]. that means we change A[i, j, k] from 0->1,or 1->0. (x1<=i<=x2,y1<=j<=y2,z1<=k<=z2).
0: “Query” operation we want to get the value of A[i, j, k].
Input
Multi-cases.
First line contains N and M, M lines follow indicating the operation below.
Each operation contains an X, the type of operation. 1: “Not” operation and 0: “Query” operation.
If X is 1, following x1, y1, z1, x2, y2, z2.
If X is 0, following x, y, z.
First line contains N and M, M lines follow indicating the operation below.
Each operation contains an X, the type of operation. 1: “Not” operation and 0: “Query” operation.
If X is 1, following x1, y1, z1, x2, y2, z2.
If X is 0, following x, y, z.
Output
For each query output A[x, y, z] in one line. (1<=n<=100 sum of m <=10000)
Sample Input
2 5 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 2 2 2 0 1 1 1 0 2 2 2
Sample Output
1 0 1
棰樼洰鍒嗘瀽 :
鏇存柊鍖洪棿, 鏌ヨ涓涓偣, 涓夌淮綰挎鏍?鐩存帴鏃犺.........鏁版嵁涓嶆槸寰堝己, 鎵浠?鐩存帴鏆村姏灝卞彲浠ヨ繃, 榪囧畬鍙戠幇鑷繁鐨勬椂闂?鍦?00MS 宸﹀彸, 鐪嬩簡涓媟ank,
灝廇 姒滈..... 鏃墮棿绔熺劧鎵?2MS
....鏋滅劧榪樻槸瑕佺敤涓夌淮鏍戠姸鏁扮粍鏉ュ姞閫熷晩 . 涓嶈繃涓鐩存病鐞嗚В 鐢?鏍戠姸鏁扮粍 瑙ｅ喅榪欓鐨勬濊礬, 浠婃棭涓婄粓浜庢槑鐧戒簡
.
鐢諱釜鍥懼厛 :
濂戒簡 , 鐪嬩唬鐮?
鏍戠姸鏁扮粍浠ｇ爜 :
/*
Mail to : miyubai@gamil.com
My Blog : www.baiyun.me
Link : http://www.cnblogs.com/MiYu || http://m.shnenglu.com/MiYu
Author By : MiYu
Test : 1
Complier : g++ mingw32-3.4.2
Program : HDU_3584
Doc Name : Cube
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
inline bool scan_d(int &num) //鏁存暟杈撳叆
{
char in;bool IsN=false;
in=getchar();
if(in==EOF) return false;
while(in!='-'&&(in<'0'||in>'9')) in=getchar();
if(in=='-'){ IsN=true;num=0;}
else num=in-'0';
while(in=getchar(),in>='0'&&in<='9'){
num*=10,num+=in-'0';
}
if(IsN) num=-num;
return true;
}
const int MAXN = 105;
int mat[MAXN][MAXN][MAXN];
int low[MAXN];
int i, j, k;
void setLow () {
for ( i = 1; i <= MAXN; ++ i ) low[i] = i & (- i);
}
void modify ( int x, int y, int z ) {
for ( i = x; i <= MAXN; i += low[i] ) {
for ( j = y; j <= MAXN; j += low[j] ) {
for ( k = z; k <= MAXN; k += low[k] )
mat[i][j][k] ^= 1;
}
}
}
int query ( int x, int y, int z ) {
int sum = 0;
for ( i = x; i > 0; i ^= low[i] ) {
for ( j = y; j > 0; j ^= low[j] ) {
for ( k = z; k > 0; k ^= low[k] )
sum += mat[i][j][k];
}
}
return sum & 1;
}
int main ()
{
setLow ();
int N, M;
while ( scan_d ( N ) && scan_d ( M ) ) {
memset ( mat, 0, sizeof ( mat ) );
while ( M -- ) {
int x1,y1,z1,x2,y2,z2;
int s; scan_d ( s );
switch ( s ) {
case 1:
scan_d ( x1 ); scan_d ( y1 ); scan_d ( z1 );
scan_d ( x2 ); scan_d ( y2 ); scan_d ( z2 );
modify ( x1,y1,z1 ); modify ( x1,y1,z2+1 );
modify ( x2+1,y1,z1 ); modify ( x2+1,y1,z2+1 );
modify ( x1,y2+1,z1 ); modify ( x2+1,y2+1,z1 );
modify ( x2+1,y2+1,z2+1 ); modify ( x1,y2+1,z2+1 );
break;
case 0:
scan_d ( x1 ); scan_d ( y1 ); scan_d ( z1 );
printf ( "%d\n", query ( x1,y1,z1 ) );
}
}
}
return 0;
}
鏆村姏浠ｇ爜 :
/*
Mail to : miyubai@gamil.com
My Blog : www.baiyun.me
Link : http://www.cnblogs.com/MiYu || http://m.shnenglu.com/MiYu
Author By : MiYu
Test : 1
Complier : g++ mingw32-3.4.2
Program :
Doc Name :
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
struct node {
int x1, x2, y1, y2, z1, z2;
}cube[10010];
int main()
{
int N, M;
int x, y, z;
while ( scanf ( "%d%d",&N,&M ) != EOF ) {
int cnt = 0;
while ( M -- )
{
int ask;
scanf ( "%d", &ask );
switch ( ask ) {
case 1:
scanf ( "%d%d%d%d%d%d", &cube[cnt].x1,&cube[cnt].y1,
&cube[cnt].z1,&cube[cnt].x2,
&cube[cnt].y2,&cube[cnt].z2 );
++ cnt;
break;
case 0:
scanf ( "%d%d%d", &x, &y, &z);
int count = 0;
for ( int i = 0; i != cnt; ++ i )
if ( cube[i].x1 <= x && x <= cube[i].x2 &&
cube[i].y1 <= y && y <= cube[i].y2 &&
cube[i].z1 <= z && z <= cube[i].z2 )
count ^= 1;;
puts ( count ? "1" : "0" );
}
}
}
return 0;
}
]]>
http://acm.hdu.edu.cn/showproblem.php?pid=1016
棰樼洰鎻忚堪:
Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6150 Accepted Submission(s): 2745
Problem Description
A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.
Note: the number of first circle should always be 1.
Note: the number of first circle should always be 1.
Input
n (0 < n < 20).
Output
The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.
You are to write a program that completes above process.
Print a blank line after each case.
You are to write a program that completes above process.
Print a blank line after each case.
Sample Input
6 8
Sample Output
Case 1: 1 4 3 2 5 6 1 6 5 2 3 4 Case 2: 1 2 3 8 5 6 7 4 1 2 5 8 3 4 7 6 1 4 7 6 5 8 3 2 1 6 7 4 3 8 5 2棰樼洰鍒嗘瀽:鍏稿瀷鐨?DFS 棰樼洰, 涓嶉渶瑕?浠涔堝壀鏋? 鐩存帴 絀蜂婦 + 鍥炴函 灝監K浜? 涓嶈繃鍊煎緱涓鎻愮殑鏄?榪欓杈撳嚭寰圔T, 涓鑸殑 鍓嶅悗 杈撳嚭 鍥炶濺 , 絎竴涓洖杞︾敤 if( n == 1 ) 鍥炶濺; 鏉ュ仛PE浜嗗ソ鍑犳, 鏈鍚庣洿鎺ュ湪紼嬪簭鏈鍚?/span> 杈撳嚭2涓洖杞︾绔熺劧灝盇浜? YM鍟?..........浠ｇ爜濡備笅:/* MiYu鍘熷垱, 杞笘璇鋒敞鏄?: 杞澆鑷?______________鐧界櫧銇眿 http://m.shnenglu.com/MiYu Author By : MiYu Test : Program : */ #include <iostream> using namespace std; bool prim[25]; int res[25]; bool hash[25]; int N; void setPrim () { memset ( prim, 0, sizeof ( prim ) ); prim[2] = prim[3] = prim[5] = prim[7] = prim[11] = prim[13] = prim[17] = prim[19] = prim[23] = true; } bool DFS ( int num , int n ) { res[n] = num; if ( n > N ) { return false; } if ( n == N - 1 ) { for ( int i = 2; i <= N; ++ i ) { if ( prim[num + i] && prim[ i + 1 ] && !hash[i] ) { res[n+1] = i; for ( int i = 1; i <= N; ++ i ) printf ( i == 1 ? "%d" : " %d",res[i] ); putchar ( '\n' ); } } } for ( int i = 2; i <= N; ++ i ) { if ( prim[ num + i ] && !hash[i] ) { hash[i] = true; DFS ( i, n + 1 ); hash[i] = false; } } return false; } int main () { setPrim (); int ca = 1; while ( cin >> N ) { sizeof ( hash, 0 , sizeof ( hash ) ); printf ( "Case %d:\n",ca++ ); hash[1] = true; DFS ( 1, 1 ); putchar ( '\n' ); } return 0; }
]]>
棰樼洰鍦板潃:
http://acm.hdu.edu.cn/showproblem.php?pid=2079
棰樼洰鎻忚堪:
Describtion
鍙堝埌浜嗛夎鐨勬椂闂翠簡錛寈hd鐪嬬潃閫夎琛ㄥ彂鍛嗭紝涓轟簡鎯寵涓嬩竴瀛︽湡濂借繃鐐癸紝浠栨兂鐭ラ亾瀛涓鍒嗗叡鏈夊灝戠粍鍚堛備綘鏉ュ府甯粬鍚с傦紙xhd璁や負涓鏍峰鍒嗙殑璇炬病鍖哄埆錛?br>
Input
杈撳叆鏁版嵁鐨勭涓琛屾槸涓涓暟鎹甌錛岃〃紺烘湁T緇勬暟鎹?br>姣忕粍鏁版嵁鐨勭涓琛屾槸涓や釜鏁存暟n(1 <= n <= 40)錛宬(1 <= k <= 8)銆?br>鎺ョ潃鏈塳琛岋紝姣忚鏈変袱涓暣鏁癮(1 <= a <= 8),b(1 <= b <= 10)錛岃〃紺哄鍒嗕負a鐨勮鏈塨闂ㄣ?br>
Output
瀵逛簬姣忕粍杈撳叆鏁版嵁錛岃緭鍑轟竴涓暣鏁幫紝琛ㄧず瀛涓鍒嗙殑緇勫悎鏁般?/span>
鍙堝埌浜嗛夎鐨勬椂闂翠簡錛寈hd鐪嬬潃閫夎琛ㄥ彂鍛嗭紝涓轟簡鎯寵涓嬩竴瀛︽湡濂借繃鐐癸紝浠栨兂鐭ラ亾瀛涓鍒嗗叡鏈夊灝戠粍鍚堛備綘鏉ュ府甯粬鍚с傦紙xhd璁や負涓鏍峰鍒嗙殑璇炬病鍖哄埆錛?br>
Input
杈撳叆鏁版嵁鐨勭涓琛屾槸涓涓暟鎹甌錛岃〃紺烘湁T緇勬暟鎹?br>姣忕粍鏁版嵁鐨勭涓琛屾槸涓や釜鏁存暟n(1 <= n <= 40)錛宬(1 <= k <= 8)銆?br>鎺ョ潃鏈塳琛岋紝姣忚鏈変袱涓暣鏁癮(1 <= a <= 8),b(1 <= b <= 10)錛岃〃紺哄鍒嗕負a鐨勮鏈塨闂ㄣ?br>
Output
瀵逛簬姣忕粍杈撳叆鏁版嵁錛岃緭鍑轟竴涓暣鏁幫紝琛ㄧず瀛涓鍒嗙殑緇勫悎鏁般?/span>
寰堢畝鍗曠殑棰樼洰, 鐪嬩笂鍘誨氨鏄洿鎺ユ瘝鍑芥暟浜? 鑰屼笖榪樻瘮杈冩爣鍑?
浠ｇ爜濡備笅:
//MiYu鍘熷垱, 杞笘璇鋒敞鏄?: 杞澆鑷?______________鐧界櫧銇眿
#include <iostream>
using namespace std;
int score[9];
int amt[9];
int num1[41];
int num2[41];
int main ()
{
int T;
while ( cin >> T )
{
while ( T -- )
{
int N,K;
cin >> N >> K;
for ( int i = 1; i<= K; ++ i )
{
cin >> score[i] >> amt[i];
}
for ( int i = 0; i <= N; ++ i )
{
num1[i] = num2[i] = 0;
}
num1[0] = 1;
for ( int i = 1; i <= K; ++ i )
{
for ( int j = 0; j <= N; ++ j )
{
for ( int k = 0; k <= amt[i] && k * score[i] + j <= N; ++ k )
{
num2[ k * score[i] + j ] += num1[j];
}
}
for ( int j = 0; j <= N; ++ j )
{
num1[j] = num2[j];
num2[j] = 0;
}
}
cout << num1[N] << endl;
}
}
return 0;
}
#include <iostream>
using namespace std;
int score[9];
int amt[9];
int num1[41];
int num2[41];
int main ()
{
int T;
while ( cin >> T )
{
while ( T -- )
{
int N,K;
cin >> N >> K;
for ( int i = 1; i<= K; ++ i )
{
cin >> score[i] >> amt[i];
}
for ( int i = 0; i <= N; ++ i )
{
num1[i] = num2[i] = 0;
}
num1[0] = 1;
for ( int i = 1; i <= K; ++ i )
{
for ( int j = 0; j <= N; ++ j )
{
for ( int k = 0; k <= amt[i] && k * score[i] + j <= N; ++ k )
{
num2[ k * score[i] + j ] += num1[j];
}
}
for ( int j = 0; j <= N; ++ j )
{
num1[j] = num2[j];
num2[j] = 0;
}
}
cout << num1[N] << endl;
}
}
return 0;
}
榪欓亾棰樼殑鏁版嵁鏄瘮杈冨皬鐨? 鎵浠ュ彲浠ョ洿鎺ユ毚鍔汚C, 榪欐牱鎯寵搗鏉? 鏆村姏浼間箮鏇村姞綆鍗曠洿瑙? 鎵浠ヤ互鍚庡仛棰樼殑鏃跺欏簲璇ュ厛鍒嗘瀽棰樼洰鏁版嵁鐨勮妯? 浠ュ厤嫻垂姣旇禌鏃墮棿, 褰撶劧,騫沖父緇冧範鏃朵笉寤鴻鏆村姏.
鏆村姏浠ｇ爜: ( 闈炴湰浜烘墍鍐? 鎹HDU涓?鍙互0k 0ms榪? 娌¤瘯榪?nbsp;)
#include <stdio.h>
int main(void)
{
int c[9], k, n, i;
int count;
int t[9], a, b;
int total = 40;
scanf("%d", &k);
while (k-- && scanf("%d%d", &total, &n))
{
for (count = i = 0; i < 9; t[i++] = 0);
for (i = 0; i < n; i++)
{
scanf("%d%d", &a, &b);
t[a] = b;
}
for (c[8] = 0; c[8] <= t[8] && c[8] * 8 <= total; c[8]++)
for (c[7] = 0; c[7] <= t[7] && c[8] * 8 + c[7] * 7 <= total; c[7]++)
for (c[6] = 0; c[6] <= t[6] && c[8] * 8 + c[7] * 7 + c[6] * 6 <= total; c[6]++)
for (c[5] = 0; c[5] <= t[5] && c[8] * 8 + c[7] * 7 + c[6] * 6 + c[5] * 5 <= total; c[5]++)
for (c[4] = 0; c[4] <= t[4] && c[8] * 8 + c[7] * 7 + c[6] * 6 + c[5] * 5 + c[4] * 4 <= total; c[4]++)
for (c[3] = 0; c[3] <= t[3] && c[8] * 8 + c[7] * 7 + c[6] * 6 + c[5] * 5 + c[4] * 4 + c[3] * 3 <= total; c[3]++)
for (c[2] = 0; c[2] <= t[2] && c[8] * 8 + c[7] * 7 + c[6] * 6 + c[5] * 5 + c[4] * 4 + c[3] * 3 + c[2] * 2 <= total; c[2]++)
{
c[1] = total - (c[8] * 8 + c[7] * 7 + c[6] * 6 + c[5] * 5 + c[4] * 4 + c[3] * 3 + c[2] * 2);
if (c[1] >= 0 && c[1] <= t[1]) count++;
}
printf("%d\n", count);
}
return 0;
}
int main(void)
{
int c[9], k, n, i;
int count;
int t[9], a, b;
int total = 40;
scanf("%d", &k);
while (k-- && scanf("%d%d", &total, &n))
{
for (count = i = 0; i < 9; t[i++] = 0);
for (i = 0; i < n; i++)
{
scanf("%d%d", &a, &b);
t[a] = b;
}
for (c[8] = 0; c[8] <= t[8] && c[8] * 8 <= total; c[8]++)
for (c[7] = 0; c[7] <= t[7] && c[8] * 8 + c[7] * 7 <= total; c[7]++)
for (c[6] = 0; c[6] <= t[6] && c[8] * 8 + c[7] * 7 + c[6] * 6 <= total; c[6]++)
for (c[5] = 0; c[5] <= t[5] && c[8] * 8 + c[7] * 7 + c[6] * 6 + c[5] * 5 <= total; c[5]++)
for (c[4] = 0; c[4] <= t[4] && c[8] * 8 + c[7] * 7 + c[6] * 6 + c[5] * 5 + c[4] * 4 <= total; c[4]++)
for (c[3] = 0; c[3] <= t[3] && c[8] * 8 + c[7] * 7 + c[6] * 6 + c[5] * 5 + c[4] * 4 + c[3] * 3 <= total; c[3]++)
for (c[2] = 0; c[2] <= t[2] && c[8] * 8 + c[7] * 7 + c[6] * 6 + c[5] * 5 + c[4] * 4 + c[3] * 3 + c[2] * 2 <= total; c[2]++)
{
c[1] = total - (c[8] * 8 + c[7] * 7 + c[6] * 6 + c[5] * 5 + c[4] * 4 + c[3] * 3 + c[2] * 2);
if (c[1] >= 0 && c[1] <= t[1]) count++;
}
printf("%d\n", count);
}
return 0;
}
]]>
棰樼洰鍦板潃:
http://acm.hdu.edu.cn/showproblem.php?pid=1164
棰樼洰鎻忚堪:
鎶婁竴涓暟鎷嗗垎鎴愮礌鏁扮殑涔樼Н.
绱犳暟鐨勬按棰?, 鍙鎶婄礌鏁版彁鍙栧嚭鏉? 鏆村姏灝卞彲浠ヤ簡.
浠ｇ爜濡備笅 :
//MiYu鍘熷垱, 杞笘璇鋒敞鏄?: 杞澆鑷?______________鐧界櫧銇眿
#include <iostream>
using namespace std;
int num[70001];
int main ()
{
for ( int i = 2; i <= 70000; ++ i )
{
if ( !num[i] )
for ( int j = 2; i * j <= 70000; ++ j )
{
num[i*j] = 1;
}
}
int N;
while ( cin >> N )
{
int n = N;
int f = 1;
while ( n != 1 )
{
int i = 2;
for ( ; i <= 65535; ++ i )
{
if ( !num[i] && n % i == 0 )
{
if ( f )
{
cout << i;
f = 0;
}
else
{
cout << "*" << i;
}
break;
}
}
n /= i;
}
cout << endl;
}
// getchar();
return 0;
}
#include <iostream>
using namespace std;
int num[70001];
int main ()
{
for ( int i = 2; i <= 70000; ++ i )
{
if ( !num[i] )
for ( int j = 2; i * j <= 70000; ++ j )
{
num[i*j] = 1;
}
}
int N;
while ( cin >> N )
{
int n = N;
int f = 1;
while ( n != 1 )
{
int i = 2;
for ( ; i <= 65535; ++ i )
{
if ( !num[i] && n % i == 0 )
{
if ( f )
{
cout << i;
f = 0;
}
else
{
cout << "*" << i;
}
break;
}
}
n /= i;
}
cout << endl;
}
// getchar();
return 0;
}
]]>
棰樼洰鍦板潃 :
http://acm.hdu.edu.cn/showproblem.php?pid=2069
鐩墠姣忔棩騫沖潎WA嬈℃暟 20嬈′互涓?..................................
鏅氫笂鍋氫竴閬撴按棰?HDOJ 2069 ACM IN HDU, 鍙堣窡涓婃鐨?TLE 鐨?-1 涓鏍? 棰樼洰鐨勬渶鍚庝竴孌墊槸鏂囦歡緇撴潫鐨勮鏄?
閮芥病鎬庝箞鑷範鍘葷湅, 緇撴灉鐩存帴灝盬A浜?............鏈鍚庡湪 鍏冨竻鐨勬彁閱掍笅, 鍘熸潵榪欓鐪熺殑寰堟按,灝辮窡鍒氬緙栫▼鏃跺仛鐨?br style="LINE-HEIGHT: normal; WORD-WRAP: break-word">100閽變拱灝忛浮,鍏浮,姣嶉浮 100鍙竴鏍? 鏄彲浠ョ洿鎺ユ毚鍔涚殑, 鍥犱負鏁版嵁涓嶅ぇ. 褰撶劧,涔熷彲浠ョ敤 姣嶅嚱鏁版潵鍋? 涓嶈繃闇瑕?br>澧炲紑杈呭姪絀洪棿 :
c1[i][j] 琛ㄧずi鍒嗛挶鐢眏涓‖甯佺粍鎴愮殑鏂規鏁?, 鐒跺悗鍦ㄥ驚鐜腑澧炲姞涓涓驚鐜?:鐢ㄤ簬淇濆瓨鍦ㄥ綋鍓嶇‖甯佹暟閲弅涓婂鍔?br> 1 - n 涓‖甯佹椂 ( k + n < =100 ) 鏂規鏁?/span>
鏆村姏浠ｇ爜:
//MiYu鍘熷垱, 杞笘璇鋒敞鏄?: 杞澆鑷?______________鐧界櫧銇眿
1 #include <iostream>
2 int main()
3 {
4 int n, d[251] = { 0 };
5 int c1, c5, c10, c25, c50;
6 for ( n = 0; n != 251; n ++ )
7 {
8 for ( c50 = 0; c50 * 50 <= n; c50 ++ )
9 {
10 for ( c25 = 0; c50 * 50 + c25 * 25 <= n; c25++ )
11 {
12 for ( c10 = 0; c50 * 50 + c25 * 25 + c10 * 10 <= n; c10++ )
13 {
14 for ( c5 = 0; c50 * 50 + c25 * 25 + c10 * 10 + c5 * 5 <= n; c5++ )
15 {
16 c1 = n - ( c50 * 50 + c25 * 25 + c10 * 10 + c5 * 5 );
17 if ( c1 + c5 + c10 + c25 + c50 <= 100 )
18 {
19 d[n]++;
20 }
21 }
22 }
23 }
24 }
25 }
26 while ( ~scanf("%d", &n) )
27 {
28 printf ( "%d\n", d[n] );
29 }
30 return 0;
31 }
32
1 #include <iostream>
2 int main()
3 {
4 int n, d[251] = { 0 };
5 int c1, c5, c10, c25, c50;
6 for ( n = 0; n != 251; n ++ )
7 {
8 for ( c50 = 0; c50 * 50 <= n; c50 ++ )
9 {
10 for ( c25 = 0; c50 * 50 + c25 * 25 <= n; c25++ )
11 {
12 for ( c10 = 0; c50 * 50 + c25 * 25 + c10 * 10 <= n; c10++ )
13 {
14 for ( c5 = 0; c50 * 50 + c25 * 25 + c10 * 10 + c5 * 5 <= n; c5++ )
15 {
16 c1 = n - ( c50 * 50 + c25 * 25 + c10 * 10 + c5 * 5 );
17 if ( c1 + c5 + c10 + c25 + c50 <= 100 )
18 {
19 d[n]++;
20 }
21 }
22 }
23 }
24 }
25 }
26 while ( ~scanf("%d", &n) )
27 {
28 printf ( "%d\n", d[n] );
29 }
30 return 0;
31 }
32
姣嶅嚱鏁頒唬鐮?:
1 #include <iostream>
2 using namespace std;
3 int c1[251][101];
4 int c2[251][101];
5 int mon[6] = { 0, 1, 5, 10, 25, 50 };
6 int sum[251];
7 int main ()
8 {
9 c1[0][0] = 1;
10 for ( int i = 1; i <= 5; ++ i )
11 {
12 for ( int j = 0; j <= 250; ++ j )
13 {
14 for ( int k = 0; k * mon[i] + j <= 250; ++ k )
15 {
16 for ( int p = 0; p + k <= 100; ++ p )
17 {
18 c2[ j + k * mon[i] ][p + k] += c1[j][p];
19 }
20 }
21 }
22 for ( int j = 0; j != 251; ++ j )
23 {
24 for ( int k = 0; k != 101; ++ k )
25 {
26 c1[j][k] = c2[j][k];
27 c2[j][k] = 0;
28 }
29 }
30 }
31 for ( int j = 0; j != 251; ++ j )
32 {
33 for ( int i = 0; i != 101; ++ i )
34 {
35 sum[j] += c1[j][i] ;
36 }
37 }
38 int N;
39 while ( cin >> N )
40 {
41
42 cout << sum[N] << endl;
43 }
44 return 0;
45 }
46
2 using namespace std;
3 int c1[251][101];
4 int c2[251][101];
5 int mon[6] = { 0, 1, 5, 10, 25, 50 };
6 int sum[251];
7 int main ()
8 {
9 c1[0][0] = 1;
10 for ( int i = 1; i <= 5; ++ i )
11 {
12 for ( int j = 0; j <= 250; ++ j )
13 {
14 for ( int k = 0; k * mon[i] + j <= 250; ++ k )
15 {
16 for ( int p = 0; p + k <= 100; ++ p )
17 {
18 c2[ j + k * mon[i] ][p + k] += c1[j][p];
19 }
20 }
21 }
22 for ( int j = 0; j != 251; ++ j )
23 {
24 for ( int k = 0; k != 101; ++ k )
25 {
26 c1[j][k] = c2[j][k];
27 c2[j][k] = 0;
28 }
29 }
30 }
31 for ( int j = 0; j != 251; ++ j )
32 {
33 for ( int i = 0; i != 101; ++ i )
34 {
35 sum[j] += c1[j][i] ;
36 }
37 }
38 int N;
39 while ( cin >> N )
40 {
41
42 cout << sum[N] << endl;
43 }
44 return 0;
45 }
46
]]>