lightoj 1213 - Fantasy of a Summation 【数学计数】

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题目链接：lightoj 1213 - Fantasy of a Summation

题意：求解

#include <stdio.h>int cases, caseno;int n, K, MOD;int A[1001];int main() {    scanf("%d", &cases);    while( cases-- ) {        scanf("%d %d %d", &n, &K, &MOD);        int i, i1, i2, i3, ... , iK;        for( i = 0; i < n; i++ ) scanf("%d", &A[i]);        int res = 0;        for( i1 = 0; i1 < n; i1++ ) {            for( i2 = 0; i2 < n; i2++ ) {                for( i3 = 0; i3 < n; i3++ ) {                    ...                    for( iK = 0; iK < n; iK++ ) {                        res = ( res + A[i1] + A[i2] + ... + A[iK] ) % MOD;                    }                    ...                }            }        }        printf("Case %d: %d\n", ++caseno, res);    }    return 0;}

公式：
ans=k∗nk−1∗(∑ni=1a[i])。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <map>#include <stack>#define PI acos(-1.0)#define CLR(a, b) memset(a, (b), sizeof(a))#define fi first#define se second#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 1e7 + 1;const int pN = 1e6;// <= 10^7const int INF = 0x3f3f3f3f;//const int MOD = 1e9 + 7;void getmax(int &a, int b) {a = max(a, b); }void getmin(int &a, int b) {a = min(a, b); }int MOD;void add(LL &x, LL y) { x += y; x %= MOD; }LL a[1010];LL Pow(LL a, int n) {    LL ans = 1;    while(n) {        if(n & 1)            ans = ans * a % MOD;        a = a * a % MOD;        n >>= 1;    }    return ans;}int main(){    int t, kcase = 1; scanf("%d", &t);    while(t--)    {        int n, k; scanf("%d%d%d", &n, &k, &MOD);        LL sum = 0;        for(int i = 0; i < n; i++) {            scanf("%lld", &a[i]);            add(sum, a[i]);        }        printf("Case %d: ", kcase++);        if(k == 1) {            printf("%lld\n", sum);        }        else {            LL ans = 1LL * k * Pow(n, k-1) % MOD * sum % MOD;            printf("%lld\n", ans);        }    }    return 0;}

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