int i,j,k=0,m=0; for(i=0;i<2;i++) { for(j=0;j<3;j++) k++; k-=j; } m=i+j; cout<<"k="<<k<<",m="<<m;
时间: 2024-03-31 21:38:13 浏览: 94
这段代码的输出结果应该是"k=3,m=5"。这是因为,首先定义了四个变量i、j、k、m,并初始化k和m的值为0。然后进入一个双重for循环,每次循环k都会加1,直到循环结束,k的值为6。接着,k会减去j的值,由于j的值在内循环结束后为3,所以k的值会减去3,变成3。内循环结束后,i的值为2,j的值为3,所以m的值为5。最后输出"k=3,m=5"。
相关问题
对以下c++代码进行改编但不改变代码准确性:#include <iostream> #include <cstdio> #include <cmath> #include <cstring> using namespace std; typedef long long LL; const int N = 105, M = 2505, L = 20; const LL INF = 1e18; int n, m, K, l; LL d[N][N], w[N][N], g[L][N][N], f[N], t[N]; struct E{ int u, v, w; } e[M]; int main() { memset(g, 0x3f, sizeof g); scanf("%d%d%d", &n, &m, &K); l = log2(K); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) if (i != j) d[i][j] = INF; for (int i = 1; i <= m; i++) { scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w); d[e[i].u][e[i].v] = min(d[e[i].u][e[i].v], (LL)e[i].w); } for (int k = 1; k <= n; k++) for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) d[i][j] = min(d[i][j], d[i][k] + d[k][j]); for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { w[i][j] = d[i][j]; for (int k = 1; k <= m; k++) w[i][j] = min(w[i][j], d[i][e[k].u] - e[k].w + d[e[k].v][j]); g[0][i][j] = w[i][j]; } } for (int c = 1; c <= l; c++) for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) for (int k = 1; k <= n; k++) g[c][i][j] = min(g[c][i][j], g[c - 1][i][k] + g[c - 1][k][j]); for (int i = 1; i <= n; i++) f[i] = d[1][i]; for (int c = 0; c <= l; c++) { if (K >> c & 1) { for (int i = 1; i <= n; i++) t[i] = f[i]; memset(f, 0x3f, sizeof f); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) f[i] = min(f[i], t[j] + g[c][j][i]); } } printf("%lld\n", f[n]); return 0; }
#include <iostream>
#include <cstdio>
#include <cmath>
#include <>
using namespace std; typedef long long LL const int N = 105, M = 2505, L = 20;
const LL INF = 1e18; int n, m, K, l;
LL d[N][N], w[N][N], g[L][N][N], f[N], t[N]; struct E{
int u, v, w;
} e[M]; int main() {
memset(g, 0x3f, sizeof g);
scanf("%d%d%d", &n, &m, &K);
l = log2(K);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) if (i != j) d[i][j] = INF;
for (int i = 1; i <= m; i++) {
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
d[e[i].u][e[i].v] = min(d[e[i].u][e[i].v], (LL)e[i].w);
}
for (int k = 1; k <= n; k++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
w[i][j] = d[i][j];
for (int k = 1; k <= m; k++)
w[i][j] = min(w[i][j], d[i][e[k].u] - e[k].w + d[e[k].v][j]);
g[0][i][j] = w[i][j];
}
}
for (int c = 1; c <= l; c++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
for (int k = 1; k <= n; k++)
g[c][i][j] = min(g[c][i][j], g[c - 1][i][k] + g[c - 1][k][j]);
for (int i = 1; i <= n; i++) f[i] = d[1][i];
for (int c = 0; c <= l; c++) {
if (K >> c & 1) {
for (int i = 1; i <= n; i++) t[i] = f[i];
memset(f, 0x3f, sizeof f);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) f[i] = min(f[i], t[j] + g[c][j][i]);
}
}
printf("%lld\n", f[n]);
return 0;
}
#include <iostream> #include <cstdlib> using namespace std; int matrixChainOrder(int p[], int n, int** m, int** s) { for(int i = 1; i <= n; i++) { m[i][i] = 0; } for(int l = 2; l <= n; l++) { for(int i = 1; i <= n-l+1; i++) { int j = i+l-1; m[i][j] = INT_MAX; for(int k = i; k <= j-1; k++) { int q = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j]; if(q < m[i][j]) { m[i][j] = q; s[i][j] = k; } } } } return m[1][n]; } int main() { int n = 3; int p[] = {8, 1, 8}; int** m = new int*[n+1]; int** s = new int*[n+1]; for(int i = 1; i <= n; i++) { m[i] = new int[n+1]; s[i] = new int[n+1]; } int minMultiplications = matrixChainOrder(p, n, m, s); cout << "The minimum number of multiplications is: " << minMultiplications << endl; cout << "The solutions to the subproblems are:" << endl; for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { cout << "(" << m[i][j] << ", " << s[i][j] << ") "; } cout << endl; } for(int i = 1; i <= n; i++) { delete[] m[i]; delete[] s[i]; } delete[] m; delete[] s; return 0; }请在此代码基础上稍加修改,避免输出的矩阵元素值异常大
这段代码存在一个问题,就是在未初始化数组 `m` 和 `s` 的值时,直接对其进行赋值操作,在这种情况下,它们的值是不确定的,可能会导致输出异常大的情况。因此,我们需要在对数组进行赋值操作之前,先将它们的值初始化为 0。
修改后的代码如下所示:
```cpp
#include <iostream>
#include <cstdlib>
using namespace std;
int matrixChainOrder(int p[], int n, int** m, int** s) {
// 初始化数组 m 和 s
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
m[i][j] = 0;
s[i][j] = 0;
}
}
for(int i = 1; i <= n; i++) {
m[i][i] = 0;
}
for(int l = 2; l <= n; l++) {
for(int i = 1; i <= n-l+1; i++) {
int j = i+l-1;
m[i][j] = INT_MAX;
for(int k = i; k <= j-1; k++) {
int q = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j];
if(q < m[i][j]) {
m[i][j] = q;
s[i][j] = k;
}
}
}
}
return m[1][n];
}
int main() {
int n = 3;
int p[] = {8, 1, 8};
int** m = new int*[n+1];
int** s = new int*[n+1];
for(int i = 1; i <= n; i++) {
m[i] = new int[n+1];
s[i] = new int[n+1];
}
int minMultiplications = matrixChainOrder(p, n, m, s);
cout << "The minimum number of multiplications is: " << minMultiplications << endl;
cout << "The solutions to the subproblems are:" << endl;
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
cout << "(" << m[i][j] << ", " << s[i][j] << ") ";
}
cout << endl;
}
for(int i = 1; i <= n; i++) {
delete[] m[i];
delete[] s[i];
}
delete[] m;
delete[] s;
return 0;
}
```
修改后的代码会对数组 `m` 和 `s` 进行初始化,初始化后再进行赋值操作,从而避免输出异常大的矩阵元素值。
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