【DP】 HDOJ 5230 ZCC loves hacking

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dp[i][j] 代表用了i个数，组成j的方案数，那么dp[i][j]可以由dp[i-1][j-i]递推而来，相当于前面i-1个数加1，最后加个1，也可以由dp[i][j-i]递推而来，相当于i个数每个数加1.。。

#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 40005#define eps 1e-12#define mod 998244353#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//#pragma comment(linker, "/STACK:102400000,102400000")#define pii pair<int, int> 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;}// headLL res[maxn];LL dp[2][maxn];void init(){int now = 0, pre;dp[now][0] = 1;for(int i = 1; i <= 450; i++) {now = now ^ 1;pre = now ^ 1;for(int j = 0; j < maxn; j++) res[j] = (res[j] + dp[pre][j]) % mod;for(int j = 0; j < i; j++) dp[now][j] = 0;for(int j = i; j < maxn; j++) dp[now][j] = (dp[pre][j - i] + dp[now][j - i]) % mod;}for(int i = 1; i < maxn; i++) res[i] = (res[i] + res[i-1]) % mod;}void work(){int c, l, r, n;scanf("%d%d%d%d", &n, &c, &l, &r);l -= c, r -= c;LL ans = res[r];if(l) ans = ((ans - res[l-1]) % mod + mod) % mod;printf("%lld\n", ans);}int main(){int _;init();scanf("%d", &_);while(_--) work();return 0;}

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