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You have N dices; each of them has K faces numbered from 1 to K. Now you have arranged the N dices in a line. You can rotate/flip any dice if you want. How many ways you can set the top faces such that the summation of all the top faces equals S?
Now you are given N, K, S; you have to calculate the total number of ways.
Input starts with an integer T (≤ 25), denoting the number of test cases.
Each case contains three integers: N (1 ≤ N ≤ 1000), K (1 ≤ K ≤ 1000) and S (0 ≤ S ≤ 15000).
For each case print the case number and the result modulo 100000007.
Sample Input |
Output for Sample Input |
|
5 1 6 3 2 9 8 500 6 1000 800 800 10000 2 100 10 |
Case 1: 1 Case 2: 7 Case 3: 57286574 Case 4: 72413502 Case 5: 9 |
///1085422276 #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<queue> #include<cmath> #include<map> #include<bitset> #include<set> #include<vector> using namespace std ; typedef long long ll; #define mem(a) memset(a,0,sizeof(a)) #define meminf(a) memset(a,127,sizeof(a)); #define memfy(a) memset(a,-1,sizeof(a)); #define TS printf("111111\n"); #define FOR(i,a,b) for( int i=a;i<=b;i++) #define FORJ(i,a,b) for(int i=a;i>=b;i--) #define READ(a,b,c) scanf("%d%d%d",&a,&b,&c) #define mod 100000007 #define inf 100000000 inline ll read() { ll x=0,f=1; char ch=getchar(); while(ch<‘0‘||ch>‘9‘) { if(ch==‘-‘)f=-1; ch=getchar(); } while(ch>=‘0‘&&ch<=‘9‘) { x=x*10+ch-‘0‘; ch=getchar(); } return x*f; } //**************************************** #define maxn 15000+5 ll dp[2][maxn],sum[maxn]; int n,k,s; int main() { int T=read(); int oo=1; while(T--) { scanf("%d%d%d",&n,&k,&s); int last=0,now=1; mem(dp); dp[now][0]=1; for(int i=1;i<=n;i++) { swap(last,now); mem(dp[now]); sum[0]=dp[last][0]; for(int j=1;j<=s;j++) { sum[j]=(sum[j-1]+dp[last][j])%mod; } for(int j=1;j<=s;j++) { if(j<=k) dp[now][j]=(sum[j-1])%mod; else dp[now][j]=(sum[j-1]-sum[j-k-1]+mod)%mod; } } printf("Case %d: ",oo++); cout<<dp[now][s]<<endl;; } return 0; }
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原文地址：http://www.cnblogs.com/zxhl/p/4908941.html
