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【DP】 HDOJ 5185 Equation

2015-03-08 16:47 357 查看
先枚举最大的数。。。然后对剩下的数做一次完全背包就可以了。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 100005
#define eps 1e-7
//#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
//LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
//head

LL dp[maxn];
int n;
LL mod;

void read()
{
scanf("%d%lld", &n, &mod);
memset(dp, 0, sizeof dp);
}

void work()
{
dp[0] = 1;
LL ans = 0;
for(int i = 1; i * (i + 1) / 2 <= n; i++) {
int sum = n - i * (i + 1) / 2;
for(int j = i; j <= n; j++) dp[j] = (dp[j] + dp[j-i]) % mod;
ans = (ans + dp[sum]) % mod;
}
printf("%lld\n", ans);
}

int main()
{
int _, __ = 0;
while(scanf("%d", &_)!=EOF) {
while(_--) {
read();
printf("Case #%d: ", ++__);
work();
}
}

return 0;
}
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