题目链接
https://www.nowcoder.com/acm/contest/85/D
思路 因为数据范围较小 ，所以 可以直接 一个一个乘
AC代码
#include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <cmath> #include <climits> #include <ctime> #include <iostream> #include <algorithm> #include <deque> #include <vector> #include <queue> #include <string> #include <map> #include <stack> #include <set> #include <numeric> #include <sstream> #include <iomanip> #include <limits> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair <int, int> pii; typedef pair <ll, ll> pll; const double PI = 3.14159265358979323846264338327; const double E = 2.718281828459; const double eps = 1e-6; const int INF = 0x3f3f3f3f; const int maxn = 1e3 + 5; const int MOD = 1e9 + 7; int ans[maxn]; struct Node { int x, y; }q[maxn]; int main() { memset(q, 0, sizeof(q)); memset(ans, 0, sizeof(ans)); int n, m, num, count = 0; scanf("%d%d", &n, &m); for (int i = 0; i <= n; i++) { scanf("%d", &num); if (num) { q[count].x = num; q[count].y = i; count++; } } for (int i = 0; i <= m; i++) { scanf("%d", &num); if (num) { for (int j = 0; j < count; j++) ans[i + q[j].y] += num * q[j].x; } } int len = n + m + 1; for (int i = 0; i < len; i++) { if (i) printf(" "); printf("%d", ans[i]); } printf("\n"); }转载于:https://www.cnblogs.com/Dup4/p/9433252.html
