HDU1398 Square Coins(生成函数)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13791 Accepted Submission(s):
9493
have square shapes but also their values are square numbers. Coins with values
of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins,
9-credit coins, ..., and 289-credit coins, are available in Silverland.
There are four combinations of coins to pay ten credits:
ten
1-credit coins,
one 4-credit coin and six 1-credit coins,
two 4-credit
coins and two 1-credit coins, and
one 9-credit coin and one 1-credit coin.
Your mission is to count the number of ways to pay a given amount using
coins of Silverland.
meaning an amount to be paid, followed by a line containing a zero. You may
assume that all the amounts are positive and less than 300.
single integer representing the number of combinations of coins should be
output. No other characters should appear in the output.
10
30
0
4
27
#include<cstdio>
#include<cstring>
#include<algorithm>
const int MAXN = ;
inline int read() {
char c = getchar(); int x = , f = ;
while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
while(c >= '' && c <= '') x = x * + c - '', c = getchar();
return x * f;
}
int cur[MAXN], nxt[MAXN], now[MAXN];
int main() {
int N;
for(int i = ; i <= ; i++)
now[i] = i * i;
while(scanf("%d", &N) && N) {
memset(cur, , sizeof(cur));
for(int i = ; i <= N; i++) cur[i] = ;
for(int i = ; now[i] <= N; i++) {
for(int j = ; j <= N; j++)
for(int k = ; j + now[i] * k <= N; k++)
nxt[j + k * now[i]] += cur[j];
memcpy(cur, nxt, sizeof(nxt));
memset(nxt, , sizeof(nxt));
}
printf("%d\n", cur[N]);
}
return ;
}
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